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torisob [31]
3 years ago
9

Can someone help me please?

Mathematics
1 answer:
Kisachek [45]3 years ago
4 0
B is the answer





B is the answer
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Rewrite 7x + 4 = x – 2 as a system. y = –(7x + 3) y = –(x – 2) y = (7x + 4) y = –(x – 2) y = (7x + 4) y = x – 2
Slav-nsk [51]

Answer: \left \{ {{y=7x+4} \atop {y=x-2}} \right.  (LAST OPTION)

Step-by-step explanation:

You have the following expression given in the problem:

7x+4=x-2

If the expression on the left side of the equation is equal to the expression on the right side of the equation, and the problem asks to rewrite it as a system, you can conclude that the first equation is:

y=7x+4

And the second equation is:

y=x-2

Therefore, you can set up the following system of equations:

\left \{ {{y=7x+4} \atop {y=x-2}} \right.

8 0
3 years ago
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If the spinner is spun 60 times, how many times can you expects the spinner to land on 2​
Masja [62]
I’m pretty sure the answer 30
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%
SVETLANKA909090 [29]

\huge \boxed{\mathbb{QUESTION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)

The remaining elements of the product matrix are found in the same way.

\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)

Simplify each element by multiplying the individual terms.

\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)

Now, sum each element of the matrix.

\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}

7 0
3 years ago
Find the value of x and y. SHOW YOUR WORK
Leona [35]

Answer:

x=21, y= 35

Step-by-step explanation:

3x+7=5x-35

(subtract seven from -35)

3x=5x-42

(subtract 5 from 3)

-2x=-42

(-42/-2= positive number)

x=21

y+75=3y+5

(subtract 75 from 5)

y=3y-70

(subtract 3y from y)

-2y=-70

y=35

6 0
3 years ago
Could use some help with these. They r confusing!!
Gala2k [10]

Step-by-step explanation:

The top part is a cone.  It's volume is:

V = ⅓ π r² h

where r is the radius (half the diameter) and h is the height.

The bottom part is a cylinder.  It's volume is:

V = π r² h

where r is the radius (half the diameter) and h is the height.

The radius of the cone is 36/2 = 18 ft, and the height is 12 ft.  So the volume is:

V = ⅓ π (18 ft)² (12 ft)

V = 4,071.5 ft³

The radius of the cylinder is 36/2 = 18 ft, and the height is 39 ft.  So the volume is:

V = π (18 ft)² (39 ft)

V = 39,697.2 ft³

Therefore, the total volume is:

4,071.5 ft³ + 39,697.2 ft³ ≈ 43,769 ft³

4 0
3 years ago
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