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3 years ago

= Knowledge Check

Mathematics

1 answer:

lbvjy [14]3 years ago

3 0

Answer:

Slope is -3.

Step-by-step explanation:

The equation to get the slope is y2-y1/x2-x1. So, plug in the points to the equation. -7-8/7-3 = -15/5 = -3/1 = -3.

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velikii [3]

Answer:o

Step-by-step explanation:

3 0

3 years ago

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Ms. Gartland bought x number of shirts for the new members of her chorus. The cost for x number of shirts including $3.99 shippi

nirvana33 [79]

Answer: The answer would be

D - 12.25x + 3.99 = 77.49

Step-by-step explanation:

If each shirt costs $12.25, that would be 12.25 times “x” amount of shirts. Then, you add the $3.99 shipping amount. Which all equals 77.49.

8 0

3 years ago

Read 2 more answers

(the stake is 14ft)

Kaylis [27]

Answer:

it‘s about

32.08 ft (if its not an answer choice try 32 or anywhere near that number)

Step-by-step explanation:

SO this is a simple example of triangles

We would need to find the heigh of the flagpole which is the leg of the triangle

The other side (14ft) is the 2nd leg of the triangle

And we know the hypotenuse is 35 ft

So we put that in Pythagorean Theroux and solve for the first missing leg

a^2+14^2=35^2

a^2+196=1225

a^2=1029

√1029 is about 32.08

So its about 32.08

5 0

3 years ago

UPS charges $8 flat fee and $0.20 for each pound. FedEx charges $3 flat fee and

Natasha_Volkova [10]

Answer:

50 lbs

Step-by-step explanation:

UPS: 8 + .20p

FedEx: 3 + .30p

8 + .20p = 3 + .30p

8 = 3 + .10p

5 = .10p

p = 50

4 0

3 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em> $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

<em>Camera 2nd has to cover the maximum angle</em>, i.e. $78.70^\circ$ .

6 0

3 years ago