All categories

3 years ago

6. which solid did the net form

Mathematics

1 answer:

In-s [12.5K]3 years ago

5 0

Answer:

Hexagonal Pyramid

You might be interested in

Please help me with this problem.

mariarad [96]

Answer:

$x^{\frac{8}{3}}y^{-\frac{5}{6}$

Step-by-step explanation:

Remember the rule for converting radicals and exponents into fractions. The power is the numerator of the fraction, and the nth root is the denominator of the fraction. Also, remember that dividing by an exponent means the exponent will be negative.

$\frac{\sqrt[3]{x^{8} } }{\sqrt[6]{y^{5} } } \\\\\frac{x^{\frac{8}{3} } }{y^{\frac{5}{6} } } \\\\x^{\frac{8}{3}}y^{-\frac{5}{6}$

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=x%20%5Cdiv%202%20%2B%20x%20%5Cdiv%203%20%3D%201%20" id="TexFormula1" title="x \div 2 + x \div

blsea [12.9K]

Answer:

Exact form : x=6/5

Decimal form : x=1.2

Mixed Number form : x=1 1/5

Step-by-step explanation: Solve for x by simplifying both sides of he equation, then isolating the variable.

Hope this helps you out! ☺

4 0

3 years ago

Hakeem subtracted 8 from a number, then multiplied the difference

Inessa [10]

Answer:

Step-by-step explanation:

(×-8)5=20

(5×-40)=20

5×(-40+40)=(20+40)

5×=60

5×÷5=60÷5

×=12

3 0

3 years ago

Read 2 more answers

Domain and range? Typically I understand domain and range but these graphs are vague in detail. If someone could help that'd be

brilliants [131]

What do you mean? Oh oh oh I used to do this, what grade you it?

6 0

3 years ago

I just need no. a please help me to prove this.

OleMash [197]

Answer: see proof below

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

Given: A + B + C = π and cos A = cos B · cos C

scratchwork:

A + B + C = π

A = π - (B + C)

cos A = cos [π - (B + C)] Apply cos

= - cos (B + C) Simplify

= -(cos B · cos C - sin B · sin C) Sum Identity

= sin B · sin C - cos B · cos C Simplify

cos B · cos C = sin B · sin C - cos B · cos C Substitution

2cos B · cos C = sin B · sin C Addition

$2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}$ Division

2 = tan B · tan C

$\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}$

<u>Proof LHS → RHS</u>

Given: A + B + C = π

Subtraction: A = π - (B + C)

Apply tan: tan A = tan(π - (B + C))

Simplify: = - tan (B + C)

$\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)$

Substitution: = -(tan B + tan C)/(1 - 2)

Simplify: = -(tan B + tan C)/-1

= tan B + tan C

LHS = RHS: tan B + tan C = tan B + tan C $\checkmark$

5 0

3 years ago