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3 years ago

Solve this I can’t seem to get it

Mathematics

1 answer:

Serjik [45]3 years ago

8 0

Answer:

288

Step-by-step explanation:

First, you have to solve the equations inside the parentheses.

(2x3) x (3x2) x (2x4)

6 x 6 x 8

Pemdas: M: Left to Right

6 x 6 = 36

36 x 8 = 288

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Evaluate a + b for a = 2 and b = 3

Vikki [24]

Answer:

Step-by-step explanation:

2+3+5

5 0

3 years ago

The junior and senior students at Mathville High School are going to present an exciting musical entitled, "Math, What is it Goo

xeze [42]

Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

$\large\boxed{\large\boxed{161}}$

Explanation:

Let's put the information in a table step-by step.

(number of remaining students)

Juniors Seniors

Condition

Initially J S

15 seniors left S - 15

Twice juniors as seniors 2(S - 15)

3/4 of the juniors left 1/4×2(S - 15)
1/3 of seniors left 2/3×(S - 15)

At the end, there were 8 more seniors than juniors:

2/3×(S - 15) - 1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

$2/3\times (S - 15) - 1/4\times 2(S - 15) = 8$

Mutilply all by 12:

$8(S - 15)-6(S - 15)=96$

Distribution property:

$8S-120-6S-90=96$

Addtion property of equalities:

$8S-6S=96+120+90$

Add like terms:

$2S=306$

Division property of equalities:

$S=306/2=153$

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

$2/3\times (S-15)+1/4\times 2(S-15)$

Just substitute S with 153 fo find the number of students that remained to perfom in the musical:

$2/3\times (153-15)+1/4\times 2(153-15)\\ \\ 2/3(138)+1/2(138)$

$161$

5 0

3 years ago

A box contains 6 red pens and 4 blue pens. Cory randomly picks a pen from

Monica [59]

Answer:

Hay probabilidad

Step-by-step explanation:

porque hay mas boligrafos rojos que azules

4 0

2 years ago

-y^2+x^3-4x^3-6y^2+5-2-x^3

Alex

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${\texttt{{Hello}}}$

$\gray{\boxed{{\textsf{trouble}}}}$

$-y ^ { 2 } +x ^ { 3 } -4x ^ { 3 } -6y ^ { 2 } +5-2-x ^ { 3 } \\$

$\green{\boxed{{\textsf{Development}}}}$

$-y ^ { 2 } +x ^ { 3 } -4x ^ { 3 } -6y ^ { 2 } +5-2-x ^ { 3 } \\$

$-y^{2}-3x^{3}-6y^{2}+5-2-x^{3} \\$

$-y^{2}-3x^{3}-6y^{2}+3-x^{3} \\$

$-y^{2}-4x^{3}-6y^{2}+3 \\$

$-7y^{2}-4x^{3}+3 \\$

$\red{\boxed{{\textsf{Answer}}}}$

$-7y^{2}-4x^{3}+3 \\$

$\red{\texttt{{\textsf{XxXDxX}}}}$

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6 0

2 years ago

How to solve this question?

Arisa [49]

F(g(x)) where you see an 'x' in f(x) put a g:

f(g(x)) = 1/(g-7), now step by step: g - 7 = 6/x+7-7=6/x, and 1/(g-7)=1/(6/x)=x/6!

I let the second one for you. Sorry if you disagree!

4 0

3 years ago