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3 years ago

I don’t understand this problem

Mathematics

1 answer:

GuDViN [60]3 years ago

4 0

z^2 - z - 72

z^2 - 9z + 8z - 72

<u>Step 1</u>: 72 = 9 * 8 and when we subtract -9+8 = -1 and that's how I went from z^2 - z - 72 to z^2 - 9z + 8z - 72.

<u>Step 2</u>: Taking out common value from first and second; and third and fourth.

z^2 - 9z = z(z - 9)

8z - 72 = 8 (z - 9)

z^2 - 9x + 8x - 72

z(z-9) + 8(z-9)

(z-9) (z+8)

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Answer:

Step-by-step explanation:

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almond37 [142]

For A B and C, you just plug in the given number

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4 years ago

Find x for each of these triangles please

notka56 [123]

Answer:

8) 10°

9) X can not be determined.

10) 10°

11) 12 units.

Step-by-step explanation:

8) The three sides of the given triangle are equal. Hence, the triangle is an equilateral triangle, hence each angle will be 60°.

So, 6x = 60°

⇒ x = 10°.

9) The three angles of the given triangle are equal. Hence, the triangle is an equilateral triangle, hence each side will be equal.

So, 6x - 5 = 6x

So, x can not be determined from this equation.

10) Δ KLM is equilateral, so, KN will bisect ∠ MKL. So, ∠ NKL = 30°

Hence, 3x = 30°

⇒ x = 10°

11) In Δ XYZ, XZ = XY , so, ∠ Z = ∠ Y.

Again, ∠ X is given to be 60°.

Therefore, each angle is 60°.

So, the triangle XYZ is equilateral, and each side will be equal.

So, 3x + 8 = 4x - 4

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3 years ago

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

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3 years ago