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2 years ago

What does 2b+9 when b=3 like the hell

Mathematics

2 answers:

BabaBlast [244]2 years ago

5 0

You plug 3 into B so 2(3) + 9 = 15

beks73 [17]2 years ago

3 0

So let’s start from the beginning. 2 is by B, so that means we have to MULTIPLY 2 by whatever B equals. B equals to 3, so we have to multiply 2 by 3, which equals 6. Then, we add 6 and 9, which is 15. Your final answer is 15. Hope that helps!

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10 plus 10 divided and rounded to next 20

BartSMP [9]

10.5 I looked it up OKAY

7 0

2 years ago

Which had the greater absolute value <br> 19 or -28

nignag [31]

Answer:

-28 has the greater absolute value.

Step-by-step explanation:

19's absolute value is 19

|19|=19

but -28's absolute value is 28

|-28|=28

and 28 is greater than 19

19<28

hope u understood and it helps

if any doubt, don't hesitate to ask in the comments

and ur welcm

8 0

3 years ago

Solve for P<br><br> I am so confused

Jlenok [28]

Answer:

3q/2

Step-by-step explanation:

3p+3q=p

2p=-3q

p=-3q/2

4 0

2 years ago

Read 2 more answers

Suppose y varies directly with x. if y=-3 when x=5 find y when x=-1

Scrat [10]

$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x

&& y={{ k }}x
\end{array}
\\ \quad \\
\textit{we know that, when}
\begin{cases}
y=-3\\
x=5
\end{cases}\implies y=kx\implies (-3)=k(5)$

solve for "k", to find the "constant of variation",
then plug it back in the y = kx, to get the equation
now
what's is y when x = -1?
well, just plug that in the equation with the found "k" value,
to get "y" :)

3 0

3 years ago

Read 2 more answers

Two mathematical questions attached below.

BartSMP [9]

Answer:

See below

Step-by-step explanation:

$9.( {m}^{3} {n}^{5} )^{ \frac{1}{4} } \\ = m^{\frac{3}{4}}n^{\frac{5}{4}}\\ \\ 10. \sqrt[5]{ \sqrt[4]{x} } \\ = \sqrt[5]{ {x}^{ \frac{1}{4} } } \\ = {( {x}^{ \frac{1}{4} } )}^{ \frac{1}{5} } \\ = {x}^{ \frac{1}{4} \times \frac{1}{5} } \\ = {x}^{ \frac{1}{20} } \\ \\ \sqrt[5]{ \sqrt[3]{ {a}^{2} } } \\ = \sqrt[5]{ {a}^{ \frac{2}{3} } } \\ = {( {a}^{ \frac{2}{3} } )}^{ \frac{1}{5} } \\ = {a}^{ \frac{2}{3} \times \frac{1}{5} } \\ = {a}^{ \frac{2}{15} }$

4 0

2 years ago