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3 years ago

20 - b/4 = 23 what is the variable number

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

3 0

Answer:

$b=-12$

Step-by-step explanation:

$20-\frac{b}{4} =23$

→ Minus 20 from both sides

$-\frac{b}{4} =3$

→ Multiply both sides by -4

$b=-12$

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600,000 1/10 of what? Please help!!

igor_vitrenko [27]

600000 is 1/10 of "x", that means x = 10/10 or a whole, what is "x"?

$\bf \begin{array}{ccll}
amount&fraction\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
600000&\frac{1}{10}\\\\
x&\frac{10}{10}
\end{array}\implies \cfrac{600000}{x}=\cfrac{\frac{1}{10}}{\frac{10}{10}}\implies \cfrac{600000}{x}=\cfrac{1}{10}
\\\\\\
\cfrac{600000\cdot 10}{1}=x$

8 0

3 years ago

Which of the following is the sum of the slopes of the line 3x+y=1 and a line perpendicular to this line? A 0 B 13 C −83 D −6

laiz [17]

Answer:

-8/3

Step-by-step explanation:

First find the slope of the line

3x+y = 1

Solve for y

y = -3x+1

This is in slope intercept form

y = mx+b where m is the slope

The slope is -3

The slopes of perpendicular lines multiply to -1

m* -3 = -1

m = 1/3

The line perpendicular has a slope of 1 / (3) = 1/3

The sum is -3 + 1/3

-9/2 + 1/3 = -8/3

3 0

3 years ago

Please and don’t just answer for the points, it’s rude

Artyom0805 [142]

Answer:

7a ( 1,1)

7b ∅ or no solution

Step-by-step explanation:

The solution to the system is where the two functions intersect

For 7a. The parabola and the line intersect at (1,1)

For 7b The parabola and the line do not intersect so there is no solution

3 0

2 years ago

Read 2 more answers

Is perpendicular to OS. is a_____line.

Step2247 [10]

I would answer the problem but first upload the diagram

6 0

3 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

7 0

2 years ago