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1 year ago

Given that ABC=DEF, what is the measure of D?

Mathematics

2 answers:

Andru [333]1 year ago

7 0

B:47 is the correct answer

ki77a [65]1 year ago

5 0

B. 47°
39 + 94 + x = 180
133 + x = 180
x = 47

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Which is greater than 4.026?

schepotkina [342]

Answer:

4.020

4.030

4.021

4.006

Step-by-step explanation:

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3 years ago

The masses of two similar objects are 24kg and 81 kg respectfully. If the surface area of the larger one is 540cm2, find the sur

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

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2 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

Find the volume of a right circular cone that has a height of 11.7 m and a base with a

yulyashka [42]

Answer:

volume of the cone = 247.98 cubic meter

Step-by-step explanation:

Height of the cone= 11.7 m

Circumference= 9 m

Radius= 9/2= 4.5 m

Volume of the cone;

= $\pi r^2\frac{h}{3}$

As, $\pi =\frac{22}{7}= 3.14$

Putting the values in the formula:

Volume of the cone is:

$=3.14*(4.5*4.5)*\frac{11.7}{3} \\\\=3.14*(4.5*4.5)*3.9\\\\=3.14*20.25*3.9\\\\=3.14*78.98\\\\=247.98m^3$

The volume of the cone is 247.98 cubic meter

8 0

2 years ago

How to solve for vertex form<br> y = -x^2 + 4x – 1

Nastasia [14]

Answer: y= -(x-2)^2+3

6 0

2 years ago