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3 years ago

Solve the quadratic equation:

Mathematics

1 answer:

Umnica [9.8K]3 years ago

7 0

Answer:

Step-by-step explanation:

$x^2+7x+10=0\qquad\implies\quad a=1\,,\ b=7\,,\ c=10\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-7\pm\sqrt{7^2-4(1)(10)}}{2(1)}=\dfrac{-7\pm\sqrt{49-40}}{2}=\dfrac{-7\pm\sqrt9}{2}\\\\ x_1=\dfrac{-7+3}2=\dfrac{-4}2=-2\,,\qquad x_1=\dfrac{-7-3}2=\dfrac{-10}2=-5$

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What's the answer to this?

beks73 [17]

Answer:

Q' = (-4, -2)

R' = (4, -2)

S' = (4, 2)

T' = (-4, 2)

Step-by-step explanation:

First, we can create a matrix to scale this rectangle by putting all the x coordinates in the top row and all the y coordinates in the bottom row and multiply it by 4.

Our initial matrix to multiply:

$4\left[\begin{array}{cccc}-1&1&1&-1\\-2&-2&-1&-1\end{array}\right]$

Moved to the origin:

$4\left[\begin{array}{cccc}-1&1&1&-1\\-0.5&-0.5&0.5&0.5\end{array}\right]$

Multiplied by four:

$\left[\begin{array}{cccc}-4&4&4&-4\\-2&-2&2&2\end{array}\right]$

This gives us the points of

Q' = (-4, -2)

R' = (4, -2)

S' = (4, 2)

T' = (-4, 2)

which are our answers.

3 0

2 years ago

PLEASE HURRY WILL MARK BRAINLEYST IF YOUR CORRECT!!!! so you will get +50 points

horrorfan [7]

Area of the figure = 112 in²

Solution:

The given figure is splitted into two shapes.

One shape is square.

Each side of square are equal.

Side = 10 inch

Area of the square = side × side

= 10 × 10

= 100

Area of the square = 100 in²

The other shape is triangle.

Base of the triangle = 10 in – 6 in = 4 inch

Height of the triangle = 16 in – 10 in = 6 inch

Area of the triangle = $\frac{1}{2}\times \text{base}\times \text{height }$

$$=\frac{1}{2}\times4\times6$

= 12

Area of the triangle = 12 in²

Area of the figure = Area of the square + Area of the triangle

= 100 in² + 12 in²

Area of the figure = 112 in²

Therefore, area of the given figure is 112 in².

7 0

3 years ago

Read 2 more answers

likoan [24]

It's easy to show that $7\tan(4x)$ is strictly increasing on $x\in\left[0,\frac\pi8\right]$ . This means

$M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3$

and

$m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7$

Then the integral is bounded by

$\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)$