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3 years ago

In need of some help please.

Mathematics

1 answer:

Hatshy [7]3 years ago

8 0

By the Pythagorean theorem
50^2 = x^2 +(x +34)^2
2500 = 2x^2 +68x + 1156
x^2 +34x -672 = 0
(x -14)(x +48) = 0
x = 14 or -48

The distance from the wall to the base of the ladder is 14 ft.

_____
7-24-25 is a Pythagorean triple. The dimensions here are double those values. It can be handy to know a few of the Pythagorean triples, as that can let you write down the answers to problems without having to go through the equations.

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What is the area of the logo?

Ludmilka [50]

Answer:

$the \: area \: of \: the \: logo \: is \: {91cm}^{2}$

Step-by-step explanation:

To determine the area of the logo you have to calculate the area of the triangle and the square that comform it and then add the four areas.

Area of the square.

To calculate the area of the square you have to calculate the square of one of its sides, following the formula:

$a = {a}^{2}$

Where "a" is the length of one of its side.

The side length of the square is a=7cm, so its area will be:

$asquare = {7}^{2} \\ asquare = {49cm}^{2}$

Area of the triangles.

The three triangles are equal, they have a base equal to the side of the square, i.e. with a length of 7cm, and their height is h=4cm.

To calculate the area of one triangle, you have to multiply its base by its height and divide by 2, following the formula:

$a = \frac{b.h}{2} \\ a = \frac{7.4}{2} \\ a = \frac{28}{2} \\ a = {14cm}^{2}$

The area calculated the correspond to one triangle, since all triangles are congruent, you have to multiply the said area by 3 to determine the area of three figures:

$atriangles = 3a \\ atriangles = {42cm}^{2}$

Now that the area of all shape are calculated, you have to add them to determine the area of the logo:

$alogo = asquare + atriangle \\ alogo = 49 + 42 \\ alogo = {91cm}^{2}$

4 0

3 years ago

Read 2 more answers

Help me with 16,17,18,and 19 please simpl

mixas84 [53]

Answer:

Ques 16)

We have to simplify the expression:

$\dfrac{t^2}{t^2+3t-18}-(\dfrac{5t}{t^2+3t-18}-\dfrac{t-3}{t^2+3t-18})\\ \\=\dfrac{t^2}{t^2+3t-18}-(\dfrac{4t+3}{t^2+3t-18})\\ \\=\dfrac{t^2-4t-3}{t^2+3t+18}$

Ques 17)

$\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{2w^2-9w+4}{(2w^2+9w-5)(w^2-w-12)}\\ \\=\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{(2w-1)(w-4)}{(2w-1)(w+5)(w+3)(w-4)}\\\\=\dfrac{3w^2+7w-7}{(w+3)(w+5)}+\dfrac{1}{(w+3)(w+5)}\\\\=\dfrac{3w^2+7w-7+1}{(w+3)(w+5)}\\\\=\dfrac{(3w-2)(w+3)}{(w+5)(w+3)}\\\\=\dfrac{3w-2}{w+5}$

Ques 18)

Let the blank space be denoted by the quantity 'x'.

$\dfrac{x}{12a^2+8a}+\dfrac{15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\ \\\dfrac{x+15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a(3a+2)}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a}=7a\\\\x+15a^2=28a^2\\\\x=28a^2-15a^2\\\\x=13a^2$

Ques 19)

Let the missing quantity be denoted by 'x'.

$\dfrac{p^2+7p+2}{p^2+5p-14}-\dfrac{x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\ \\\dfrac{p^2+7p+2-x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\\\\dfrac{p^2+7p+2-x}{(p-2)(p+7)}=\dfrac{p-1}{p-2}\\\\p^2+7p+2-x=(p-1)(p+7)\\\\p^2+7p+2-x=p^2+6p-7\\\\x=p+9$

7 0

2 years ago

I'm stuck on a question I'm helping my brother with. We need to find out if y = -x squared is linear or not.

Anna71 [15]

Y=-x^2 is non linear due to the x being squared

4 0

2 years ago

How much does 1 yard equal??

zhenek [66]

1 yd equals 3 feet. 1 yd equals 36 inches.

5 0

3 years ago

Hello, could you please explain this to me?

Viefleur [7K]

Find the value of r(q(4)), so first you need to find the value of q(4).

q(4), this means that x = 4, so substitute/plug it into the equation to find the value of q(x) when x = 4:

q(x) = -2x - 1 Plug in 4 into "x" since x = 4

q(4) = -2(4) - 1

q(4) = -8 - 1

q(4) = -9

Now that you know the value of q(4), you can find the value of r(x) when x = q(4)

r(x) = 2x² + 1

r(q(4)) = 2(q(4))² + 1 Plug in -9 into "q(4)" since q(4) = -9

r(q(4)) = 2(-9)² + 1

r(q(4)) = 2(81) + 1

r(q(4)) = 163 163 is the value of r(q(4))

5 0

3 years ago