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3 years ago

The answer that I found is the incorrect answer

Mathematics

1 answer:

vovikov84 [41]3 years ago

7 0

I can’t see the image can you post it please or just tell me what the problem is

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Please Hurry ...Which expression is equivalent to

swat32

Answer:

$\huge\boxed{\sf \frac{160rs^5}{t^6}}$

Step-by-step explanation:

$\sf 5r^6t^4 ( \frac{4r^3s^tt^4}{2r^4st^6} ) ^5$

Using rule of exponents $\sf a^m/a^n = a^{m-n}$

$\sf 5r^6t^4 ( 2 r^{3-4} s^{2-1}t^{4-6})^5\\5r^6t^4(2r^{-1}st^{-2})^5\\5r^6t^4 * 32 r^{-5}s^5t^{-10}$

Using rule of exponents $\sf a^m*a^n = a^{m+n}$

$\sf 160 r^{6-5}s^5t^{4-10}$

$\sf 160 rs^5 t^{-6}$

To equalize the negative sign, we'll move t to the denominator

$\sf \frac{160rs^5}{t^6}$

8 0

3 years ago

a circular stained glass window frame with a diameter 20 inches is divided into 8 congruent find the area of 3 of the 8 panes of

ahrayia [7]

Answer:

it would be 45

Step-by-step explanation:

6 0

3 years ago

(10-4)+20 divided by 10?

Luba_88 [7]

Answer:

Step-by-step explanation:

(10-4)+20/20

6+20/10

6+2

6 0

3 years ago

Determine 2nd term and rth term of a.p.whose 6 th term is 12 and 8 th term is 22

Vanyuwa [196]

Answer:

see explanation

Step-by-step explanation:

The n th term of an arithmetic progression is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given $a_{6}$ = 12 and $a_{8}$ = 22, then

a₁ + 5d = 12 → (1)

a₁ + 7d = 22 → (2)

Subtract (1) from (2) term by term to eliminate a₁

2d = 10 ( divide both sides by 2 )

d = 5

Substitute d = 5 into (1) to find a₁

a₁ + 5(5) = 12

a₁ + 25 = 12 ( subtract 25 from both sides )

a₁ = - 13

Thus

$a_{2}$ = - 13 + 5 = - 8

$a_{n}$ = - 13 + 5(n - 1) = - 13 + 5n - 5 = 5n - 18 ← n th term

8 0

3 years ago

100 POINTS AND BRAINLIEST

zysi [14]

Answer:

8 square units and $\frac{40}{3}$ square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

4 0

3 years ago

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