3 years ago

Carl cut two triangles boards from the corners of a rectangular board as shown

Mathematics

2 answers:

TiliK225 [7]3 years ago

7 0

Answer:

I don't see a shaded area but all the shapes together would be 52.5 cm².

Step-by-step explanation:

5 × 10.5 = 52.5

zmey [24]3 years ago

7 0

Answer:

Step-by-step explanation:

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Fifteen times the square of a non-zero number is equal to 15 times the number

defon

15·x²=15·x so
x²=x
x·x=x with x≠0 than x must be 1
x=1

5 0

3 years ago

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What is the measurement of each side of the perimeter is 94?

sveticcg [70]

2a+1 + 3a+2 +5a-9 you would start by combining like terms
2a+3a+5a= 10a
1+2-9= -6
10a-6=94
then subtract 94-6=88
then divide 88 by 10= 8.8
a=8.8

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3 years ago

I NEED IT NOW!!! I ALSO NEED YOUR WORK OF HOW YOU GET YOUR ANSWER !! THANK YOU!

Mars2501 [29]

Answer:

Solve the rational equation by combining expressions and isolating the variable

All real numbers

Interval Notation:

−∞,∞)

Step-by-step explanation:

3 0

3 years ago

Am I correct? Please show your work so I can compare it to my work. Thank you!

Viefleur [7K]

Answer:

200

Step-by-step explanation:

C = 0.007x² − 2.8x + 100

There are two ways we can find the minimum. The first is by noticing that this is an upwards parabola, so the minimum is at the vertex. For y = ax² + bx + c, the vertex is at x = -b/(2a).

x = -(-2.8) / (2 × 0.007)

x = 200

Alternatively, we can minimize C using calculus. Take the derivative with respect to x and set to 0:

dC/dt = 0.014x − 2.8

0 = 0.014x − 2.8

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4 0

3 years ago

The face of a triangular shaped guitar has an area of 52 in.² and a height of 13 inches. Write an equation to find the length of

Neko [114]

Area of a triangle can be calculated as:

Area = 0.5 x Base x Height

Area is given to be 52 in²
Height is 13 inches.
We are to find the base b of the triangular shaped guitar.

Using the values in above formula, we get:

$52=0.5(13)(b) \\ \\ 
b= \frac{52}{0.5(13)} \\ \\ 
b=8$

Thus the base of triangular shaped guitar will be 8 inches.

5 0

3 years ago

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