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Mazyrski [523]
3 years ago
14

12+3x+12-x= I need help please

Mathematics
1 answer:
Marina86 [1]3 years ago
7 0
I hope this is helpful

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4. Lydia cut out her initial from a piece<br><br> of construction paper,
Ratling [72]

Answer:

Option A is correct.

10 square centimeters.

Step-by-step explanation:

Complete Question

The complete Question is attached in the first attached image.

Lydia cut out her initial from a piece of construction paper. How many square centimeters of construction paper are used to make Lydia's initial?

A) 10 square centimeters

B) 11 square centimeters

C) 15 square centimeters

D) 22 square centimeters

Solution

From the second attached image, it is evident that we can split the L-shaped figure into two rectangles of dimensions (3 cm by 1 cm) and (7 cm by 1 cm)

The total area of the figure is thus

(3 × 1) + (7 × 1) = 10 cm²

Hope this Helps!!!

3 0
3 years ago
jordan go's to the bank every 7 days and then Clare go's to the bank every day will bolth go to the bank
RUDIKE [14]

Answer:

on the seventh day

Step-by-step explanation:

hope this helps

8 0
3 years ago
NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z​
Roman55 [17]

Answer:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

Step-by-step explanation:

Given system of equations:

\begin{cases}y=x^2-9\\y=4x-4\end{cases}

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}

Factor the quadratic:

\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}

Apply the <u>zero-product property</u> and solve for x:

\implies x+1=0 \implies x=-1

\implies x-5=0 \implies x=5

Substitute the found values of x into the <u>second equation</u> and solve for y:

\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}

\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}

Therefore, the solutions are:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

6 0
1 year ago
Read 2 more answers
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