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3 years ago

Guys I have another problem it’s easy

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

8 0

The answer would be C. 8

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1. You go to the local Sport Grill for quarter wings day.

DedPeter [7]

Answer:

If the 20% discount is by the hour then the answer would be $11.70 I think.

0.25x78= 19.50

19.50x0.20(20%)= 3.90

19.50-3.90-3.90= 11.70

Hope this helps!

Step-by-step explanation:

8 0

3 years ago

Find the value of x. A. 60 B. 50 C. 40 D. 30

garri49 [273]

We can see that two sides of that triangle area equal

so, their corresponding angle will also be equal

now, we know that

sum of all angles in any triangle is always 180

so, we can get equation as

$x+100+x=180$

now, we can solve for x

$2x+100=180$

$2x=80$

$x=40$

so, option-C..............Answer

8 0

3 years ago

Read 2 more answers

A company logo is made up of a square and three identical triangles. What is the area of the logo? Enter your answer in the box.

garri49 [273]

Given:
square with sides measuring 7 cm.
3 triangles attached to three sides of the square. A line bisecting one triangle is measured at 4 cm.

Area of a square = s² = (7cm)² = 49 cm²

Area of a triangle = hb/2 = (4cm*7cm)/2 = 14 cm²
Area of the 3 triangles = 14 cm² x 3 = 42 cm²

Total area of the logo = 49 cm² + 42 cm² = 91 cm²

7 0

3 years ago

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In ΔABC, the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively.

Irina-Kira [14]

Given the values of the three sides of the triangle, we can apply the Cosine Law to find the angles of the triangle. Recall that for we can express the value of c through the equation below.

$c^{2} = a^{2} + b^{2} - 2abcosC$

Rearranging this equation, we can find the value ∠C as shown below.

$\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$
$C = cos^{-1} (\frac{a^{2}+b^{2}-c^{2}}{2ab})$

We can apply the same reasoning for finding the value of ∠B as shown.

$B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})$

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

$C = cos^{-1} (\frac{22.5^{2}+18^{2}-13.6^{2}}{2(22.5)(18)})$
$C \approx 37.19$

$B = cos^{-1} (\frac{22.5^{2}+13.6^{2}-18^{2}}{2(22.5)(13.6)})$
$B \approx 53.13$

Answer: ∠C = 37.19° and ∠B = 53.13°

7 0

4 years ago

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In what quadrant is the terminal side of -323°?

dexar [7]

Answer:

Quadrant 4 i believe plz tell me if im wrong

Step-by-step explanatio

6 0

3 years ago

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