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2 years ago

Help! Solve for x !! Use picture below

Mathematics

1 answer:

creativ13 [48]2 years ago

7 0

Answer:

x=7

Step-by-step explanation:

since it is the same side length,

3x+6 = 10x-43

3x-10x = -43-6

-7x = -49

x=-49/-7=7

please mark brainliest

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Eloise hung artwork on 2/5of a bulletin board.she hung math papers on 1/5of the same bulletin board.what part of the bulletin bo

erica [24]

Here, number of bulletin board = 1
Artwork = 2/5 th, & math papers = 1/5

So, it would be: 1/5 + 2/5 = 3/5

In short, Your Answer would be 3/5 or 0.60

Hope this helps!

7 0

3 years ago

I need help with 4 and 5 what are the answers?

ira [324]

Answer:

4. U = 54

5. C=80

Step-by-step explanation:

Basically you move the denominator to the other side and multiply. please take note of the signs(positve, negative)

7 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

The equation 4x2 – 24x + 4y2 + 72y = 76 is equivalent to

DerKrebs [107]

Answer:

Option 4 is correct.

The equation $4x^2 -24x + 4y^2 + 72y = 76$ is equivalent to $4(x-3)^2 + 4(y+9)^2 =436$

Step-by-step explanation:'

Given equation: $4x^2 -24x + 4y^2 + 72y = 76$

First group the terms with x and those with y;

$(4x^2-24x)+(4y^2+72y) = 76$

Next, we complete the squares.

We can do this by adding a third term such that the x terms and the y terms are perfect squares.

For this we must either add the same value on the other side of the equation or subtract the same value on the same side so that the equality is maintained.

⇒ $4(x^2-6x) +4(y+18y) = 76$

$4(x^2 -6x +3^2 -3^2) + 4(y^2 +18y +9^2 -9^2) = 76$

$4(x^2-6x + 3^2) - 36 + 4(y^2+18y +9^2) - 324 = 76$

$4(x-3)^2 + 4(y+9)^2 - 360 =76$

Add 360 on both sides we get;

$4(x-3)^2 + 4(y+9)^2 =360 +76$

Simplify:

$4(x-3)^2 + 4(y+9)^2 =436$

Therefore, the given equation is equivalent to $4(x-3)^2 + 4(y+9)^2 =436$

5 0

3 years ago

The slope of a line is 7/10 . One point on the line has coordinates (1, 2). Which ordered pair represents another point on the l

mash [69]

(x,y)

y=mx+b
m=slope
b=yintercept
slope=7/10
one point is (1,2) or x=1, then y=2
find b
y=7/10x+b
2=7/10(1)+b
2=7/10+b
minus 7/10 both sides
1 and 3/10=b
13/10=b
y=7/10x+13/10
input values for x and get values for y
one point is (0,13/10)

7 0

3 years ago