All categories

3 years ago

Please help with the first question

Mathematics

1 answer:

NemiM [27]3 years ago

8 0

Answer:

<h2>I Think Letter A</h2>

Step-by-step explanation:

Hope It Helps

You might be interested in

Solve x2 – 7x = –13.

aleksklad [387]

we have

$x^{2} -7x=-13$

Complete the square. Remember to balance the equation by adding the same constants to each side

$x^{2} -7x+3.5^{2}=-13+3.5^{2}$

$x^{2} -7x+12.25=-13+12.25$

$x^{2} -7x+12.25=-0.75$

Rewrite as perfect squares

$(x-3.5)^{2}=-0.75$

remember that

$i=\sqrt{-1}$

Square root both sides

$(x-3.5)=(+/-)\sqrt{-0.75}$

$(x-3.5)=(+/-)\sqrt{-1*0.75}$

$(x-3.5)=(+/-)\sqrt{0.75}i$

$x1=3.5+\sqrt{0.75}i$

$x2=3.5-\sqrt{0.75}i$

Simplify

$\sqrt{0.75}=\frac{\sqrt{3}}{2}$

$x1=3.5+\frac{\sqrt{3}}{2}i$

$x2=3.5-\frac{\sqrt{3}}{2}i$

therefore

<u>the answer is</u>

$x1=3.5+\frac{\sqrt{3}}{2}i$

$x2=3.5-\frac{\sqrt{3}}{2}i$

6 0

3 years ago

Read 2 more answers

Help pleaseeee!!!!!!!

mylen [45]

Help with what? I need more information

6 0

3 years ago

Find the radius of a circle with an area of 380 in².

Likurg_2 [28]

Answer:

11 inches

Step-by-step explanation:

The area of a circle can be found using:

$a=\pi r^{2}$

We already know that the area is 380, so we can substitute 380 in for a.

380= $\pi r^2$

Now, we need to find the radius. To do this, we need to get r by itself

First, divide both sides by pi

$380/\pi =\pi r^2/\pi$

120.95775675=r^2

Since r is being squared, we need to take the square root of both sides

$\sqrt{120.95775675}=\sqrt{r^2}$

10.9980796846=r

If we round to nearest whole number, our radius is 11 inches

5 0

3 years ago

Find the solutions to the system below<br><br> y=x2+6x<br><br> y=x−4

dangina [55]

both and then combine them or just find them 1 by 1

7 0

3 years ago

Suppose that the augmented matrix for a linear system has been reduced by row operations to the given row echelon form. Solve th

erastovalidia [21]

Answer:

a) x = -37 , y = -8, z = 5.

b) w = 13z-10, x = 13z-5, y= 2-z, z = free parameter.

c) v = 2z-7w-11, w = free parameter, x = -3z-4, y = 9-3z, z = free parameter

d) No solution

Step-by-step explanation:

a) $\left[\begin{array}{cccc}1&-3&4&7\\0&1&2&2\\0&0&1&5\end{array}\right]$

The following set of equations are:

z = 5

y+2z=2

x-3y+4z=7

Solving above equations:

y = -8, x = 7-4(5)+3(-8) = -37

b) $\left[\begin{array}{ccccc}1&0&8&-5&6\\0&1&4&-9&3\\0&0&1&1&2\end{array}\right]$

The following set of equations are:

y+z = 2

x+4y-9z=3

w+8y-5z=6

Solving above equations:

z = free parameter - No. of columns > No. of rows; hence, multiple solutions.

y = 2-z, x = 3+9z-4(2-z) = -5+13z, w = 6+5z-8(2-z) = 13z-10

c) $\left[\begin{array}{cccccc}1&7&-2&0&-8&-3\\0&0&1&1&6&5\\0&0&0&1&3&9\end{array}\right]$

The following set of equations are:

y + 3z = 9

x + y + 6z = 5

v + 7w -2x -8z = -3

Solving above equations:

z & w = free parameter - No. of columns > No. of rows; hence, multiple solutions.

y = 9-3z, x = -3z-4, v = -11+2z-7w

d) $\left[\begin{array}{cccc}1&-3&7&1\\0&1&4&0\\0&0&0&1\end{array}\right]$

The following set of equations are:

0 = 1

y + 4z = 0

x-3y+7z = 1

Solving above equations:

No Solution as the first equation is inconsistent

5 0

3 years ago