All categories

2 years ago

What value of c makes the equation true?

Mathematics

1 answer:

8_murik_8 [283]2 years ago

3 0

Comparing forms, you see that
4 = ∛c

The value of c is found by cubing both sides of this equation.
4³ = c = 64

You might be interested in

Which function equation is represented by the graph. a) f(x) = 50(1.5)^x B) f (x) =50(1.8)^x. C) f(x) = 50(2.8)^x D) f(x) = 50(2

bearhunter [10]

I would say the third one would be the correct or the second one

7 0

3 years ago

Read 2 more answers

A quadrilateral has two angles that measure 222° and 81°. The other two angles are in a ratio of 6:13. What are the measures of

hram777 [196]

$\sf\huge\underline{\star Solution:-}$

$\rightarrow$ <u>Ratio of two angles are 6:13.</u>

So, let the one angle be $\sf{6x}$ and another be $\sf{13x.}$

As we know that,

Sum of all interior angles of a quadrilateral = $\sf\pink{360°.}$

So let's sum up the given angles,

$\rightarrow$ $\sf{222°+81°+6x+13x \:=\: 360°}$

$\rightarrow$ $\sf{303+19x \:=\: 360°}$

$\rightarrow$ $\sf{19x \:=\: 360-303}$

$\rightarrow$ $\sf{19x \:=\: 57}$

$\rightarrow$ $\sf{x \:=\: \frac{19}{57}}$

$\rightarrow$ $\sf{x \:=\: 3}$

Hence, $\sf{x \:=\: 3}$

So, First angle $\sf{= \:6x\: = \:6×3 \:=}$ $\sf\purple{18°.}$

Second angle $\sf{= \:13x\: = \:13×3 \:=}$ $\sf\purple{39°.}$

_________________________________

Hope it helps you:)

3 0

2 years ago

5 divided by -5 is what?

mart [117]

5 divided by -5 is -1 hope this helps

5 0

3 years ago

Read 2 more answers

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes

azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

The centre of the ellipse is at the origin and the X axis is the major axis

It passes through the points (-3, 1) and (2, -2)

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

Cross multiplying we get,

9b² + a² = 1 ²× a²b²
a²b² = 9b² + a²

Multiply by 4 on both sides,

4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

Cross multiply,

4b² + 4a² = 1 × a²b²
a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
20b² = 128
b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>

8 0

3 years ago

Write three different ways for7/10

HACTEHA [7]

The only ways i thought of was 0.7, 0.70, and 0.700 .

7 0

3 years ago

Read 2 more answers