5. For which linear function below does f(-1) = 9 and f(3) = 1?

PLEASE HELP I HAVE 5 MIN TO TURN THIS IN!!!!

LUCKY_DIMON [66]

Answer:

3- False,
4- True

Step-by-step explanation:

<h3>Garden A</h3>

<u>Rectangle with sides</u>

Width ⇒ 2x - 3 feet
Length ⇒ 5(2x - 3) = 10x - 15

<u>Perimeter is</u>

2(w + l) = 2( 2x - 3 + 10x - 15) = 24x - 36

<h3>Garden B</h3>

<u>Triangle with sides</u>

5x + 5, 4x + 2, 8x - 8

<u>Perimeter is</u>

5x + 5 + 4x + 2 + 8x - 8 = 17x - 1

<h3>Statement 3</h3>

<u>The difference in perimeters is the difference is fencing:</u>

24x - 36 - (17x - 1) = 24x - 17x - 36 + 1 = 7x - 35

Since it is different from 7x - 37 the statement is FALSE

<h3>Statement 4</h3>

Garden A is surrounded by a path of 2.5 ft wide

It will have the perimeter added by 2.5*2*4 = 20 ft as each side is extended by 2*2.5 ft = 5ft

<u>So the perimeter is going to be:</u>

24x - 36 + 20 =
24x - 16

This statement is TRUE

7 0

3 years ago

Please help solve and explain this thank you

vova2212 [387]

Answer:

hi

Step-by-step explanation:

4 0

3 years ago

Choose the product. a 3 b 2 · 4ab 3 <br><br> 4a3b5<br><br> 4a3b6<br><br> 4a4b5

zvonat [6]

A³ b² 4ab³

Rearrange order:

4 a³ a b² b³

Now add up the exponents from same base:

4 a³⁺¹ b²⁺³

4 a⁴ b⁵

Final answer: 4 a⁴ b⁵

5 0

3 years ago

There's a room, the height is 12, length is 14, and width is 10. What is the total area of the 4 walls?

Scorpion4ik [409]

Answer:

Hello Everybody Love Pizza

Step-by-step explanation:

7 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

4 years ago