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3 years ago

Using the theorems and given information, which of the following proves a∥b?

Mathematics

1 answer:

stich3 [128]3 years ago

4 0

Answer:

i am sorry bro later i will answer

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Van guessed on all 8 questions of a multiple-choice quiz. Each question has 4 answer choices. What is the probability that he go

s344n2d4d5 [400]

Answer:

0.267 is the answer!!!

8 0

3 years ago

What is the value of the following expression 1.8 + 0.5(y + 6) - 2 t the power of 3 when y = 14?

Vikentia [17]

Answer:

The value of given expression is 3.8.

Step-by-step explanation:

Given:

$1.8+0.5(y+6)-2^3$

We need to find the simplified value of given expression using y =14

Solution:

Now we will substitute y =14 we get;

$1.8+0.5(y+6)-2^3\\\\1.8+0.5(14+6)-2^3$

Now using PEDMAS we will first solve the parenthesis we get;

$1.8+0.5\times20-2^3$

Now we will solve the exponent function we get;

we know that;

$2^3=8$

So we can say that;

$1.8+0.5\times20-8$

Now we will solve the multiplication operation.

$1.8+10-8$

Now we will perform addition operation.

$11.8-8$

And finally we will perform subtraction operation.

3.8

Hence The value of given expression is 3.8.

3 0

4 years ago

What is the answer to 3/5 x 10/21 in simplest form???

jeka94

6/21 I think cause you can't go down any more

3 0

4 years ago

Read 2 more answers

The figure is made up of a hemisphere and a cylinder.

Sidana [21]

Answer:

90π in^3.

Step-by-step explanation:

Total Volume = volume of cylinder + volume of the hemisphere.

= πr^2h + 1/2*4/3πr^3

= π * 3^2 * 8 + 2/3 π 3^3 ( Note the radius r = 1/2 diameter = 1/2*6 = 3)

= 72π + 18π

= 90π in^3.

3 0

3 years ago

Can someone help me solve this differentiation/tangent problem?

nirvana33 [79]

A)

$\bf g'(x)=\stackrel{product~rule}{2x\cdot f(x)+x^2\cdot f'(x)}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
g'(5)=2(5)\cdot f(5)+(5)^2\cdot f'(5)\implies g'(5)=50+500\\\\\\ g'(5)=550\\\\
-------------------------------\\\\
g(5)=(5)^2f(5)\implies g(5)=125
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}
\begin{cases}
x=5\\
y=125\\
\stackrel{m}{g'(5)}=550
\end{cases}\implies y-125=550(x-5)
\\\\\\
y-125=550x-2750\implies y=550x+400$

b)

$\bf h'(x)=\stackrel{quotient~rule}{\cfrac{f'(x)(x-6)~~-~~f(x)\cdot 1}{(x-6)^2}}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
h'(5)=\cfrac{f'(5)(5-6)~~-~~f(5)\cdot 1}{(5-6)^2}
\\\\\\
h'(5)=\cfrac{5(-1)-5}{(-1)^2}\implies h'(5)=-10\\\\
-------------------------------\\\\$

$\bf h(5)=\cfrac{f(5)}{5-6}\implies h(5)=\cfrac{5}{-1}\implies h(5)=-5
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad 
\begin{cases}
x=5\\
y=-5\\
\stackrel{m}{-10}
\end{cases}\implies y-(-5)=-10(x-5)
\\\\\\
y+5=-10x+50\implies y=-10x+45$

3 0

4 years ago