The equivalent expression is seven-fifths squared times one-third cubed. Then the correct option is D.
<h3>What is an equivalent expression?</h3>
The equivalent is the expression that is in different forms but is equal to the same value.
Five-sevenths squared times one-third raised to the power of negative three all raised to the power of negative one.
Convert the sentence into numerical form.
![\rm \rightarrow \left [ \left (\dfrac{5}{7} \right )^2 \times \left ( \dfrac{1}{3}\right )^{-3} \right ] ^{-1}](https://tex.z-dn.net/?f=%5Crm%20%5Crightarrow%20%5Cleft%20%5B%20%5Cleft%20%28%5Cdfrac%7B5%7D%7B7%7D%20%5Cright%20%29%5E2%20%5Ctimes%20%5Cleft%20%28%20%5Cdfrac%7B1%7D%7B3%7D%5Cright%20%29%5E%7B-3%7D%20%5Cright%20%5D%20%5E%7B-1%7D)
Simplify the expression, then we have
![\rm \rightarrow \left [ \left (\dfrac{7}{5} \right )^{2} \times \left ( \dfrac{1}{3}\right )^{3} \right ]](https://tex.z-dn.net/?f=%5Crm%20%5Crightarrow%20%5Cleft%20%5B%20%5Cleft%20%28%5Cdfrac%7B7%7D%7B5%7D%20%5Cright%20%29%5E%7B2%7D%20%5Ctimes%20%5Cleft%20%28%20%5Cdfrac%7B1%7D%7B3%7D%5Cright%20%29%5E%7B3%7D%20%5Cright%20%5D)
The equivalent expression is seven-fifths squared times one-third cubed. Then the correct option is D.
More about the equivalent link is given below.
brainly.com/question/889935
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Answer:
C
Step-by-step explanation:
formula: <u>Y</u><u>2</u><u> </u><u>-</u><u> </u><u>Y</u><u>1</u>
X2 - X1
-18 = X1
16 = Y1
31 = X2
40 = Y2
<u>4</u><u>0</u><u> </u><u>-</u><u> </u><u>1</u><u>6</u><u>. </u><u> </u><u> </u> = <u>2</u><u>4</u>
31 - (-18) 49
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
Answer:
126
Step-by-step explanation:
14 times 9
