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2 years ago

HELP ASSAP WITH THIS QUESTION

Mathematics

2 answers:

attashe74 [19]2 years ago

8 0

If I remember correctly judging from the 30 in the top left all you would have to is subtract 180 which is the angle of a straight line and subtract it from the 30 and it would be 150

krek1111 [17]2 years ago

4 0

That would be 180 - 30 = 150 degrees

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You decide to get an estimate on the cost of repairs for the green food truck. The mechanic explains that the cost of the repair

I am Lyosha [343]

Answer:

you save labor : 3562.50

Step-by-step explanation:

You want to calculate for labor.

labor = 3/5 (parts)

labor + parts = $9500 (total costs)

3/5 parts + parts =9500

3/5 parts + 5/5 parts = 9500

8/5 parts=9500

5/8(8/5 parts) = 5/8 (9500)

‘parts = 5937.50

labor=3562.50 (3/5 x 5937.50)

total = 9500

7 0

2 years ago

Geometry - Will give 5 stars + Thanks!

Fynjy0 [20]

Answer:

Step-by-step explanation:

We know:

The diagonals in a parallelogram divide by halves.

Therefore KG = UG and DG = CG.

We have

KG = 5y - 8, UG = 3y, DG = 4x - 7, CG = x + 2

Substitute:

5y - 8 = 3y add 8 to both sides

5y = 3y + 8 subtract 3y from both sides

2y = 8 divide both sides by 2

y = 4

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4x - 7 = x + 2 add 7 to both sides

4x = x + 9 subtract x from both sides

3x = 9 divide both sides by 3

x = 3

5 0

2 years ago

Please please help me!!

makvit [3.9K]

Answer:

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Step-by-step explanation:

6 0

2 years ago

Can somebody help me

Sergeeva-Olga [200]

$cis(x)=\cos x+i\sin x\\\\cis15^o=\cos15^o+i\sin15^o\\\\\cos15^o=\dfrac{\sqrt6+\sqrt2}{4};\ \sin15^o=\dfrac{\sqrt6-\sqrt2}{4}\\\\\text{therefore}\\\\7cis15^o=7\left(\cos15^o+i\sin15^o\right)=7\cos15^o+7i\sin15^o\\\\=7\left(\dfrac{\sqrt6+\sqrt2}{4}\right)+7\left(\dfrac{\sqrt6-\sqrt2}{4}\right)i$

6 0

2 years ago

Is w^2-9w+81 a trinomial square

Yuri [45]

No because it would be
if you knew that
ax^2+bx+c was a trinomial square, then the factored form would be
$( \sqrt{a}x+ \sqrt{c})^{2}$ or $( \sqrt{a}x- \sqrt{c})^{2}$
so
if w^2-9w+81 is perfect square, factored form is either
$( \sqrt{1}x+ \sqrt{81})^{2}$ or $( \sqrt{1}x- \sqrt{81})^{2}$ aka $( x+9)^{2}$ or $( x-9)^{2}$
expand them
(x+9)^2=x^2+18x+81, wron middle term

(x-9)^2=x^2-18x+81 wrong

so it is not

8 0

2 years ago