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babymother [125]

3 years ago

9

Are this too answer correct

1 answer:

Ede4ka [16]3 years ago

4 0

I believe so friend budddy

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Hello there ^ _ ^ <br><br> Can you help me please ? Please explain how you find the answer. Thanks!

ziro4ka [17]

It's C because for the perpendicular line, you need to construct cross points to draw it through.

- you start by drawing a random circle with F as the center, just as long as it intersects line DE
- Then you use the intersections of that circle to draw new circles starting at F.
- These new circles intersect at F and another point below DE.
- You draw the perpendicular line through F and the other point.

8 0

3 years ago

What is the slope in simplest form?

oksian1 [2.3K]

Answer:the slope is equal to zero ur welcome

Step-by-step explanation:

4 0

3 years ago

A brother and sister share out their collection of 5000 stamps in the

yaroslaw [1]

Answer:

1875

Step-by-step explanation:

After sharing in the ratio 5:3, the brother has

5000/(5+3) x 5 = 3125

Next, he shares with his friend in the ratio 3:1:1, thus

3125/(3+1+1) x 3 = 1875

7 0

3 years ago

Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

or

$x = \frac{6\pm\sqrt{36-12}}{2}$

or

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

or

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

or

$x = \frac{4\pm\sqrt{16+56}}{4}$

or

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

or

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

or

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

or

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

7 0

3 years ago

Yuki888 [10]

Answer:

28

Step-by-step explanation:

f(6) = -3 * 6 + 10 = -18 + 10 = -8

g(f(6)) = g(-8)

g(-8) = $(-8)^{2} + 3 * (-8) - 12$ = 64 - 24 - 12 = 64 - 36 = 28

6 0

3 years ago