Does 10x = 2.5x have one solution, no solutions, or infinite solutions?

The value closest to the lateral area of the cylinder is 138. 5cm^2. Option C

The formula for determining the lateral area of a cylinder is expressed as;

Lateral area = 2 πrh

Where;

Note that radius is half the diameter;

radius, r = 5. 8/ 2 = 2. 9 cm

Substitute into the formula

Lateral area = 2 ( 3. 142 × 2. 9 × 7. 6)

Lateral area = 2 ( 69. 249)

Lateral area = 138. 49 cm^2

Thus, the value closest to the lateral area of the cylinder is 138. 5cm^2. Option C

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Answer:

1st one is the right option

The value of x is 4 , 12 , - 1 resp. in eq 1 , 2 and 3 .

Finding values of unknown variable x .

Following are linear equations in one variable .

On solving them we get ,

1 )          2 ( 3 x - 5 ) = - 4 x + 30

             6 x - 10 = - 4 x + 30

              10 x =40

                 x = 4

2 )          5 x - 6 - 3 x = 18

              2 x = 18 + 6

              2 x = 24

               x = 12

3 )         - 11 - 5 x = 6 ( 5 x + 4 )

               - 11 - 5 x = 30 x + 24

                 - 35 x = 35

                x = - 1

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Answer:

3

Step-by-step explanation:

input -3 into h(x)=x^2-6

h(-3)=-3^2-6=9-6=3

Answer:

5/4 > 5/6

Step-by-step explanation:

Answer:

k = 1

Step-by-step explanation:

since 360 = 60² , a perfect square , then

360 × 1 ← is a perfect square

that is 360k with k = 1 is a perfect square

Answer:

10°C

Step-by-step explanation:

c = [tex]\frac{F - 32}{1.8}[/tex]  Plug in 50 for F and solve for C

c = [tex]\frac{50-32}{1.8}[/tex]

C = [tex]\frac{18}{1.8}[/tex] Divide

c = 10

If a high school student decides to apply four of six famous colleges, then the four colleges can be selected in as many as 15 different combinations.

As per question statement, a high school student decides to apply four of six famous colleges.

We are required to calculate the number of combinations in which the four colleges can be selected.

To solve this question, we need to know the formula of Combination which goes as [tex](nCr)=\frac{n!}{r!(n-r)!}[/tex]

, i.e., we are to select a set or "r" from a set of "n".

Here, we have to select a combination of 4 from a set of 6.

Therefore applying (nCr) formula with (n = 6) and (r = 4), we get,

[tex](6C4)=\frac{6!}{4!(6-4)!} =\frac{6!}{4!2!}=\frac{4!*5*6}{4!*2}=\frac{5*6}{2}=(5*3)=15[/tex].

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Step-by-step explanation:

2(x-1) + 5(x-9)

2x - 2 + 5x - 45

2x + 5x - 2 - 45

7x - 47

Answer:

2x -2 + 5x -45       7x -47

Step-by-step explanation:

2*x = 2x

2*-1 = -2

5*x =5x

5*-9 = -45

Answer: [tex]\boxed{y=\frac{2}{5} x-7}\ as\ an\ example[/tex]

Step-by-step explanation:

Find the slope-intercept form:

[tex]2x-5y=30\\\\-5y=-2x+30\\\\y=\frac{-2x+30}{-5} \\\\y=\frac{2}{5} x-6[/tex]

Therefore the slope is 2/5

An example of a line parallel to this line is

[tex]y=\frac{2}{5} x-7[/tex]

(A) is false. By symmetry,

[tex]\displaystyle f(x) = \int_{-x}^x (|t|_t^2) e^{-t^2} \, dt = 2 \int_0^x (t-t^2) e^{-t^2} \, dt[/tex]

where [tex]|t|=t[/tex] since [tex]x>0[/tex]. Substitute [tex]s=t^2[/tex] to get the equivalent integral,

[tex]\displaystyle f(x) = \int_0^{x^2} (1 - \sqrt s) e^{-s} \, ds[/tex]

Then

[tex]\displaystyle f(x) + g(x) = \int_0^{x^2} e^{-s} \, ds[/tex]

[tex]\displaystyle f(\sqrt{\ln(3)}) + g(\sqrt{\ln(3)}) = \int_0^{\ln(3)} e^{-s} \, ds = \frac23 \neq \frac13[/tex]

(B) is false. Note that [tex]1+\alpha x[/tex] is linear so its derivative is the constant [tex]\alpha[/tex] at every point. We then have

[tex]{\psi_1}'(\alpha) = -e^{-\alpha}+1 = \alpha \implies 1-\alpha = e^{-\alpha}[/tex]

But this has no solutions, since the left side is negative for [tex]\alpha>1[/tex] and the right side is positive for all [tex]\alpha[/tex].

(C) is true. By the same reasoning as in (B), the line [tex]2x(\psi_1(\beta)-1)[/tex] has constant derivative, [tex]2\psi_1(\beta)-2 = 2e^{-\beta+2\beta-2[/tex]. Then

[tex]{\psi_2}'(\beta) = 2\beta-2+2e^{-\beta} = 2e^{-\beta}+2\beta-2[/tex]

holds for all values of [tex]\beta[/tex].

(D) is false. We use the first derivative test. By the fundamental theorem of calculus,

[tex]\displaystyle f(x) = 2 \int_0^x (t-t^2)e^{-t^2}\,dt \implies f'(x) = 2(x-x^2)e^{-x^2}[/tex]

Solve for the critical points.

[tex]f'(x) = 0 \implies x-x^2 = 0 \implies x = 0 \text{ or } x = 1[/tex]

[tex]e^{-x^2}>0[/tex] for all [tex]x[/tex], so the sign of [tex]f'[/tex] depends on the sign of [tex]x-x^2[/tex]. It's easy to see [tex]f'>0[/tex] for [tex]x\in(0,1)[/tex] and [tex]f'<0[/tex] for [tex]x\in\left(0,\frac32\right)[/tex]

AAnswer:

AB (-1, 4) = 34^ and B = (3,-1) 249*32 = 20

Step-by-step explanation:

Answer:

[tex]AB=\sqrt{53}\:\sf units[/tex]

[tex]EF=\sqrt{349}\:\sf units[/tex]

Step-by-step explanation:

Distance between two points

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points}.[/tex]

Given:

Substitute the given points into the distance formula and solve for AB:

[tex]\begin{aligned}\implies AB & =\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\& =\sqrt{(3-(-4))^2+(-1-1)^2}\\& =\sqrt{(7)^2+(-2)^2}\\& =\sqrt{49+4}\\& = \sqrt{53}\end{aligned}[/tex]

Given:

Substitute the given points into the distance formula and solve for EF:

[tex]\begin{aligned}\implies EF & =\sqrt{(x_F-x_E)^2+(y_F-y_E)^2}\\& =\sqrt{(11-(-7))^2+(3-(-2))^2}\\& =\sqrt{18^2+5^2}\\& =\sqrt{324+25}\\& =\sqrt{349}\end{aligned}[/tex]

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The given set is (3, 4, 5, ...} is infinite set.

A set with an infinite number of elements is one that cannot be numbered. A set that has no last element is said to be endless. A set that can be put into a one-to-one correspondence with a suitable subset of itself is said to be infinite. No issue with the in-class assignment.

The stars in the clear night sky, water droplets, and the billions of cells in the human body are just a few examples of endless sets of objects that surround us. A set of natural numbers, however, serves as the best illustration of an infinite set in mathematics. There is no limit to the amount of natural numbers.

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Answer:

C.) Equally as distant.

Answer:

[tex]V=lwh\\\frac{V}{lw} = \frac{lwh}{lw} \\\frac{V}{lw} = h[/tex]

Answer:

x = 23 degrees

Because if you look at diagram x is vertical to angle FBC, and angle FEC and angle EBA are alternate interior angles so they are equal to each other, and AC is a straight line which means it equals to 180 degrees.

[tex]4x+65+x = 180[/tex]

x = 23 degrees

Answer:

1,803.268

Step-by-step explanation:

so thousandths place is three number after the decimal just see the fourth number after the decimal and if it 5 or greater round the (in this case) 8 but since the fourth number was 4 it stays the same but only three numbers

1) Yes, because each value of x maps onto only one value of y.

2) No, because the x-value of 1 maps onto 4 different y values.

Answer:

Step-by-step explanation:

You want to know if the given sets of ordered pairs represent a function.

A function is a relation that maps an input value to a single output value. On a graph, this means no two points are vertically aligned. For a table or set of ordered pairs, it means no input (x) value is repeated.

The input (x) values are {-1, 0, 1, 2} with no repeats. This relation is a function.

The input values are {1, 1, 1, 1} with repeats. This relation is not a function.

We get that the equation of the line PQ will be 2 x - 7 y + 17 = 0.

A) We know that the equation of the line is:

y - y₁ = m (x - x₁)

where m = y₂ - y₁ / x₂ - x₁

Now, we have:

P = ( x₁, y₁) = (2, 3)

Q = (x₂, y₂) = (9, 5)

We get m by substituting the values as:

m = 5 - 3 / 9 - 2 = 2 / 7

The equation then will be:

y - y₁ = 2 / 7 (x - x₁)

Substituting the values, we get that:

y - 3 = 2/7 ( x - 2)

7 y - 21 = 2 x - 4

2 x - 7 y - 4 + 21 = 0

2 x - 7 y + 17 = 0

B) When PQ will intersect x-axis, then the point will be = ( x, 0)

2 x - 7 y + 17 = 0

2 x - 7 (0) + 17 = 0

2 x = -17

x = - 17 / 2

Point = (-17 / 2 , 0)

Therefore, we get that the equation of the line PQ will be 2 x - 7 y + 17 = 0.

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Answer: 576

Step-by-step explanation:

5+3 = 8

8*72=576

Answer:576

Step-by-step explanation: just add and multiply

Answer:

15

Step-by-step explanation:

The sides add to 63, so:

[tex]11+3x+11+x+2+2x+3=63 \\ \\ 6x+27=63 \\ \\ 6x=36 \\ \\ x=6 \\ \\ \implies AB=2(6)+3=15[/tex]

The number of nails in each box is: 1300, 1400, 600, 1200, and 500

Given 5000 nails in five boxes.

Let X denote the universal set.

Then n(X) = 5000.

Let A denote the first box,

B denote the second box,

C denote the third box,

D denote the fourth box, and

E denotes the fifth box.

The first and second boxes have 2700 nails altogether.

⇒n(A ∪ B) = 2700

The second and third boxes have 2000 nails altogether.

⇒n(B ∪ C) = 2000

The third and fourth boxes have 1800 nails altogether.

⇒n(C ∪ D) = 1800

The fourth and fifth boxes have 1700 nails altogether.

⇒n(D ∪ E) = 1700

We need to find out how many nails are there in each box.

That is to find out: n(A), n(B), n(C), n(D), and n(E).

Note that all these five sets are disjoint. This means that the intersection is empty.

n(A ∪ B ∪ C ∪ D) = n(A ∪ B) + n(C ∪ D) = 2700 + 1800 = 4500

n(E) = n(X) - n(A ∪ B ∪ C ∪ D) = 5000 - 4500 = 500

n(D) = n(D ∪ E) - n(E) = 1700 - 500 = 1200

n(C) = n(C ∪ D) - n(D) = 1800 - 1200 = 600

n(B) = n(B U C) - n(C) = 2000 - 600 = 1400

n(A) = n(A U B) - n(B) = 2700 - 1400 = 1300

Therefore, the number of nails in each box is 1300, 1400, 600, 1200, and 500.

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answer and steps to get it in the picture

According to this the temperature in February = 4.5°F.

The absolute value (or modulus) of a real number x is its non-negative value regardless of its sign. For example, 5 has an absolute value of 5, and 5 has an absolute value of 5. A number's absolute value can be conceived of as its proximity from zero along the real number line.

The temperature on the same day in February has the same absolute vale as the temperature in the January, but is not the same temperature according to this the temperature in February = 4.5°F.

The absolute value is the same but not the same temperature:

|-4.5| = 4.5

So,

Its 4.5°F

According to this the temperature in February = 4.5°F.

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If

[tex]f(x) = \dfrac{2ax+b}{x-1}[/tex]

Note that

[tex]f(2) = \dfrac{4a+b}{2-1} = 4a+b[/tex]

From the given information we have

[tex]\displaystyle \lim_{x\to0} f(x) = \lim_{x\to0} \frac{2ax+b}{x-1} = \frac{0+b}{0-1} = -b = -3 \\\\ ~~~~ \implies b=3[/tex]

and

[tex]\displaystyle \lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{2ax+b}{x-1} = \lim_{x\to\infty} \frac{2a+\frac bx}{1-\frac1x} = \frac{2a+0}{1-0} = 2a = 4 \\\\ ~~~~ \implies a=2[/tex]

It follows that

[tex]f(2) = 4\cdot2+3 = 11[/tex]

as required.

Answer: She can use 4 gigabyte while remaining in her budget.

Step-by-step explanation:

You would first subtract 54  from 70 which leaves you with 16,

Then you would divide 16 by 4 to find how many gigabytes she could use which gives you 4

So she can use 4 gigabyte while remaining in her budget.

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Answer: Answer is 3

Step-by-step explanation:

let the maximum number of gigabytes of data is x

total budget is  $70

then, flat cost is $54 and $5per gigabyte

we get:

54 +5 × x= 70

5x =70 - 54

x = [tex]\frac{16}{5}[/tex]  = 3.2 ≈ 3

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Answer:

$3285.47

Step-by-step explanation:

We know the value decreased by a factor of 2500/28000 after 5 years.

This means the value decreased by a factor of (2500/28000)⅕ per year.

Therefore, after 3 years, the value is

[tex]28000(2500/28000)^{3/5} \approx 3285.47[/tex]