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3 years ago

An equation with more than one operation is called

Mathematics

1 answer:

exis [7]3 years ago

4 0

Divide subtract multiply addition

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Can someone help me?

Alex73 [517]

No i will not help you

5 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Bogdan [553]

The radical expressions and their corresponding expression in rational exponent form are

$972^\frac 15 = 3\sqrt[5]{4}$ , $448^\frac 13 = 4\sqrt[3]{7}$ , $3528^\frac 12 = 42 \sqrt 2$ and $4050^\frac 14 = 3\sqrt[4]{50}$

<h3>How to match the tiles?</h3>

We start by evaluating each expression as follows:

$972^\frac 15$

Expand 972

$972^\frac 15 = (243 * 3)^\frac 15$

Take the 5th root of 243

$972^\frac 15 = 3* (4)^\frac 15$

Rewrite as:

$972^\frac 15 = 3\sqrt[5]{4}$

Next, we have:

$448^\frac 13$

Expand 448

$448^\frac 13 = (64 * 7)^\frac 13$

Take the 3rd root of 64

$448^\frac 13 = 4 * (7)^\frac 13$

Rewrite as:

$448^\frac 13 = 4\sqrt[3]{7}$

Next, we have:

$3528^\frac 12$

Expand 3528

$3528^\frac 12 = (1764 * 2)^\frac 12$

Take the square root of 1764

$3528^\frac 12 = 42 * (2)^\frac 12$

Rewrite as:

$3528^\frac 12 = 42 \sqrt 2$

Lastly, we have:

$4050^\frac 14$

Expand 4050

$4050^\frac 14 = (81 * 50)^\frac 14$

Take the 4th root of 81

$4050^\frac 14 = 3 (50)^\frac 14$

Rewrite as:

$4050^\frac 14 = 3\sqrt[4]{50}$

Read more about radical expressions at:

brainly.com/question/8952483

#SPJ1

5 0

2 years ago

HEY FRIENDS ITS JAMIE!!! Hope ur day is good.

pav-90 [236]

THE SECOND ONE IS THE CORRECT ANSWER (B)

good luck

5 0

4 years ago

Read 2 more answers

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

4 years ago

In the model below what is x and how do I solve it

tiny-mole [99]

Answer:so that’s a interior triangle so the mean your 2 points in the inside and the equation on the outside.

X=6

Step-by-step explanation:

(8x-4)+(3x+17)=(17x-23)

Combined our alike terms on the left side of the equation. So 8x+3x and -4+17.

11x+13=17x-23

Now this is what the equation should look like after combining ur alike terms.

11x+13=17x-23

-11x. -11x.

now u take the lowest x which is 11x and mince that from 17x but u only mince if it’s a positive but is it’s a negative u add it.

13=6x-23 so if u moved the x on the right side u move non variable number to the left but if u moved the x to the left u move the non variable number to the right so in this case the non variable number would be -23 so now u add 23 on both sides.

36=6x and now u just divide the 36 by 6 and remember to always divide by the x

Now the answer is x=6

7 0

4 years ago