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3 years ago

Write 2 ratios that are equivalent to 3:4 explain how they are equivalent

1 answer:

8 0

6:8
9:12
They are equivalent because to get, for example, 6:8, we had to multiply each side of the ratio 3:4 by the same number. the same goes for 9:12.

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How do i solve a triangles area

shusha [124]

Answer:

Area = 1/2 Base x height

Step-by-step explanation:

To find the area of a triangle, multiply the base of the height and then divide by 2. The division by 2 comes from the fact that a parallelogram can be divided into two triangles.

6 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%20%7Btan%7D%5E%7B3%7D%20x%20%5C%3A%20dx" id="TexFormula1" title=" \int {tan}^{3} x \

r-ruslan [8.4K]

Answer:

$\frac{ {tan}^{2} x}{2} + \ln( |cos \: x| ) + C$

Step-by-step explanation:

$\int {tan}^{3} x \: dx$

$\int \: tan \: x \times {tan}^{2} x \: dx$

$\int \: tan \: x( {sec}^{2} x - 1) \: dx$

distribute

$\int \: tan \: x \: {sec}^{2} x - tan \: x \: dx$

$\int \: tan \: x \: {sec}^{2} x \: dx \: - \int \: tan \: x \: dx$

$\int \: tan \: x \: {sec}^{2} x \: dx \: - \int \frac{sin \: x}{cos \: x} \: dx$

First integrand

let tan x = u

du = sec²x dx

Second integrand

let cos x = z

dz = -sin x dx

$= \int u \: du \: - \int - \frac{1}{z} dz$

$= \frac{ {u}^{2} }{2} + \ln( |z| ) + C$

$= \color{red}{ \boxed{ \frac{ {tan}^{2} x}{2} + \ln( |cos \: x| ) + C}}$

7 0

3 years ago

Write the expression below as a single exponent.

ivann1987 [24]

Answer:

Step-by-step explanation:

3 0

2 years ago

What is the x-intercept on the equation y=2x-10

Mrrafil [7]

The x-intercept is the point where the graph intersects the y-axis.
At the y-axis, all points have 0 for the y-coordinate.
Let y = 0, and solve for x.

y = 2x - 10

0 = 2x - 10

10 = 2x

5 = x

x = 5

The x-intercept is 5.

5 0

4 years ago

Read 2 more answers

What is the point of maximum growth:

salantis [7]

f(h(x))= 2x -21

Step-by-step explanation:

f(x)= x^3 - 6

h(x)=\sqrt[3]{2x-15}

WE need to find f(h(x)), use composition of functions

Plug in h(x)

f(h(x))=f(\sqrt[3]{2x-15})

Now we plug in f(x) in f(x)

f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6

cube and cube root will get cancelled

f(h(x))= 2x-15 -6= 2 x-21

wait i think i did the wrong one brb

7 0

4 years ago