Can somebody answer this for me?

What is the solution to this system of equations?

Degger [83]

Answer:

Step-by-step explanation:

8 0

2 years ago

The sides of a triangle have lengths x + 3, 4(x - 13), and 3 over 4 x. The perimeter of the triangle is 3x +6. What is the lengt

nignag [31]

Answer:

shortest side = 15

Step-by-step explanation:

The question is a bit ambiguous. Do you mean (3/4) x or do you mean 3/(4x)?

I'll take it to be the first one.

(x + 3) + 4(x - 13) + (3/4)x = 3x + 6 Remove the brackets on the left

x + 3 + 4x - 52 + 0.75 x = 3x + 6 Combine

5.75x - 49 = 3x + 6 Subtract 3x from both sides.

2.75x - 49 = 6 Add 49 to both sides

2.75x = 55 Divide by 2.75

x = 55/ 2.75

x = 20

Now for the shortest side

x + 3 = 23

4(x - 13) = 4(20 - 13) = 4*7 = 28

(3/4)*20 = 15

The shortest side = 15

3 0

3 years ago

In the year 2017. Silver Creek Middle School had 288 students. In the year 2018, the school

Harman [31]

Answer:

12.5 % increase

Step-by-step explanation:

percent increase: $\frac{final-initial}{initial}$ x 100

324-288 = 36/288 = 0.125 x 100 = 12.5 %

7 0

2 years ago

Two angles are ____________________ if they are across from each other and are congruent (same measure).

dimulka [17.4K]

Answer:

Vertical Angles

8 0

2 years ago

(1-i)^2 find 4 th root

sergiy2304 [10]

By de Moivre's theorem,

$1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}$

$\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}$

where $k\in\{0,1,2,3\}$ . The fourth roots of $(1-i)^2$ are then

$k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}$

$k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}$

$k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}$

$k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}$

or more simply

$\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}$

We can go on to put these in rectangular form. Recall

$\cos^2(x) = \dfrac{1 + \cos(2x)}2$

$\sin^2(x) = \dfrac{1 - \cos(2x)}2$

Then

$\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

$\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

and the roots are equivalently

$\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}$

7 0

1 year ago