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2 years ago

Fhdryyyyyyyyyyyyetttttttttt

Mathematics

2 answers:

german2 years ago

7 0

Answer:

fjkn QLJKFnclqnmjfn,msfjv

Step-by-step explanation:

noname [10]2 years ago

4 0

Answer:

jsoanjcisnzocisjsncjhsjakznc aknxosjz

Step-by-step explanation:

hajxkfj snksksnc

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How to solve this inequality -8×+18>-22

Ksju [112]

Answer:

x < 5

Step-by-step explanation:

Step 1: Solve for x

-8x + 18 > -22

-8x + 18 - 18 > -22 - 18

-8x / -8 > -40 / -8

When dividing by negative, it flips the sign

x < 5

Answer: x < 5

4 0

2 years ago

Write the equation for the line that passes through the points (4, 5) and (6,9). *

N76 [4]

Answer:

y = 2x + 1

Step-by-step explanation:

first find slops

(9-5)/(6-4) = 4/2 = 2 = m

y = mx + b

5 = 2(2) + b

1 = b

y = 2x + 1

5 0

2 years ago

Read 2 more answers

Help

djverab [1.8K]

Answer:

x=7

Step-by-step explanation:

Find slope first

$m-\frac{y2-y1}{x2-x1}$

$m=\frac{6-5}{7-7}$

$m=\frac{1}{0}\\$

m is undefined- so it is a vertical line

x=7

6 0

2 years ago

You get a 3% commission on all sales. This month, you made a sale of $45,050 and

shtirl [24]

Answer:

$1,555.06

Step-by-step explanation:

8 0

2 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago