All categories

2 years ago

PLEASE HELP ME WITH THIS!!!! ILL GIVE BRAINLIEST

Mathematics

2 answers:

MAVERICK [17]2 years ago

8 0

Answer:

The answer is D.

Step-by-step explanation:

because the height of shap A is 10 and the height of shape B is 30, that means it was multiplied by 3.

Rina8888 [55]2 years ago

7 0

Answer:

Below.

Step-by-step explanation:

We have 6/6 the radius and 10/30 that means the height of b is 3 times greater then a therefore the answer is D.

You might be interested in

Forgot to add pictures but are these answers correct? Brainliest

nignag [31]

Answer:

i think so!

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A wire 390 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures ha

jeka57 [31]

Answer:

Step-by-step explanation:

Let x is the length of the square side.

We know: the two figures have the same area

<=> $x^{2}$ = π $r^{2}$ (r being radius)

<=> x = $\sqrt{II}$ r

perimeter square = 4x = 4*r* $\sqrt{II}$

perimeter circle = 2*r*π

<=> 390 = 4*r* $\sqrt{II}$ + 2*r*π

<=> r = 29.16

=> perimeter square = 4*29.16*π

=> perimeter circle = 58.32π

3 0

3 years ago

Read 2 more answers

Solve the system of linear equations by substitution.<br><br> y = 2x + 8<br> y = -2x

posledela

Answer:

(- 2, 4 )

Step-by-step explanation:

y = 2x + 8 → (1)

y = - 2x → (2)

substitute y = 2x + 8 into (2)

2x + 8 = - 2x ( add 2x to both sides )

4x + 8 = 0 ( subtract 8 from both sides )

4x = - 8 ( divide both sides by 4 )

x = - 2

substitute x = - 2 into (2)

y = - 2(- 2) = 4

solution is (- 2, 4 )

6 0

2 years ago

Read 2 more answers

The local softball field has a lemonade stand. The graph shows the amount of money in the cash register of the lemonade stand.

Alexxandr [17]

<h3>The y-intercept</h3>

The y-intercept is $35.

The y-intercept of a straight line graph is the point at which the graph intercepts the y-axis.

So, from the graph, the y-intercept is $35.

<h3>The meaning of y-coordinate</h3>

The y-coordinate of the y-intercept of the function represents the amount of money present at the beginning of the sale when no cups where sold.

Since the y-intercept represents the point at which the graph intercepts the y-axis and then the x- value is zero, the y-coordinate of the y-intercept of the function represents the amount of money present at the beginning of the sale when no cups where sold.

The y-coordinate of the y-intercept of the function represents the amount of money present at the beginning of the sale when no cups where sold.

<h3>The slope</h3>

The slope is 2.5

Since we are given two points on the line, the slope m of a line given two points (x₁, y₁) and (x₂, y₂) is m = (y₂ - y₁)/(x₂ - x₁).

From the graph, (x₁, y₁) = (3, 42.5) and (x₂, y₂) = (9, 57.5)

So, substituting these values into the equation for m, we have

m = (y₂ - y₁)/(x₂ - x₁)

m = (57.5 - 42.5)/(9 - 3)

m = 15.0/6

m = 2.5

The slope is 2.5

<h3>Meaning of the slope</h3>

The slope represents the rate at which the lemonade is sold

Since the y-axis represent the amount of money in the cash register and the x-axis represents the number of cups of lemonade sold, the slope represents the rate at which the lemonade is sold.

The slope represents the rate at which the lemonade is sold

Learn more about straight-line graphs here:

brainly.com/question/14808962

8 0

2 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago