All categories

3 years ago

Write 5 times x to the two thirds power in radical form.

Mathematics

1 answer:

Citrus2011 [14]3 years ago

8 0

(5 • x)^2/3

We know that the parentheses goes around 5 • x and that it also goes to the power of 2/3.

You might be interested in

What is the missing statement in the proof? Scroll down to see the entire proof.

Eva8 [605]

Answer: C. $\angle BCA\cong \angle DCB$ is the missing statement.

Explanation: Given: ΔABC with m∠ABC = 90° (view diagram)

Prove: $AB^2 + BC^2 = AC^2$

Proof: 1. Draw $BD\perp AC$ (construction)

2. $\angle ABC\cong\angle BDC$ (Angles with the same measure are congruent.)

3. $\angle BCA\cong \angle DCB$ (Reflexive Property of Congruence)

4. $\triangle ABC\sim \triangle BDC$ (AA criterion for similarity)

5. BC:DC=AC:BC (Corresponding sides of similar triangles are proportional.)

6. $BC^2= AC\times DC$ (cross multiplication)----------(1)

7. $\angle ABC\cong \angle ADB$ (Angles with the same measure are congruent)

8. $\angle BAC\cong \angle DAB$ (Reflexive Property of Congruence)

9. $\triangle ABC\sim \triangle ADB$ (AA criterion for similarity)

10. AB:AD=AC:AB⇒ $AB^2=AC.AD$ --------------(2)

equation (1) + equation (2) ⇒ $AB^2+BC^2= AC(DC+AD)$ ⇒ $AB^2+BC^2= AC^2$

6 0

3 years ago

Read 2 more answers

Unit 8 Correlation and Regression

julsineya [31]

Wait can someone answer this I need help too

7 0

3 years ago

Which number sentence is true?

IRINA_888 [86]

Answer:

D is the true answer.

Step-by-step explanation:

0.006262 is smaller than 0.062

7 0

3 years ago

Help if you can please

professor190 [17]

Answer:1.)-27

Step-by-step explanation:

6 0

3 years ago

Find the value of cosec^2 30 degree - sin^45 - sec^60

Sedbober [7]

<u>For this question, I will be writing 'csc' instead of 'cosec'</u>

<u /> $\sin 30^{\circ}=\frac{1}{2} \implies \csc 30^{\circ}=2\\\\\sin 45^{\circ}=\frac{1}{\sqrt{2}}\\\\\cos 60^{\circ}=\frac{1}{2} \implies \sec 60^{\circ}=2\\\\\\\therefore \csc^{2} 30^{\circ}-\sin^{2} 45^{\circ}-\sec^{2} 60^{\circ}=2^{2}-\left(\frac{1}{\sqrt{2}} \right)^{2}-2^{2}=\boxed{-\frac{1}{2}}$

4 0

2 years ago