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2 years ago

WILL GIVE BRAINLIEST

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

8 0

Answer:

Range=65

Step-by-step explanation:

The highest is 66 the lowest is 1 so 66-1=65

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F(x) = -2x + 5 f ( ) = 13

scZoUnD [109]

Answer:

Okay, I haven't done this a long time however I believe with this you have to solve the equation by doing substitution.

Step-by-step explanation:

so, if f (x) equals 13, you replace the x in the equation to 13.

f(x)=-2x+5

f(x)=-2(13)+5

f(x)=26+5

f(x)=31

I believe 41 would be your answer. Hope this is right and it helped!

4 0

3 years ago

Help with the this question pleaseeee

34kurt

Answer:

32x - 8

Step-by-step explanation:

A = LW

A = (8x - 2)(4)

A = 32x - 8

7 0

2 years ago

Read 2 more answers

What’s the degree of

Leno4ka [110]

The degree would be 3

7 0

3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is:

$(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})$

$(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]$

$(x,y) = (10.452, 34.531)\,[km]$

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

$\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}$

$\theta \approx 16.840^{\circ}$

And the distance from the camp is calculated by the Pythagorean Theorem:

$r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}$

$r = 36.078\,km$

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0

2 years ago

The area of a rectangular television screen is 3456 in^2. The width of that screen is 24 inches longer than the length. What is

shutvik [7]

1. The problem says that the television has a rectangular shape. So, the formula for caculate the area of a rectangle is:

A=LxW

"A" is the area of the rectangle (A=3456 inches²).
"L" is the the length of the rectangle.
"W" is the width of the rectangle.

2. The width of the screen is 24 inches longer than the length. This can be expressed as below:

W=24+L

3. Then, you must substitute W=24+L into the formula A=LxW:

A=LxW
 3456=L(24+L)
3456=24L+L²

4. The quadratic equation is:

L²+24L-3456=0

5. When you solve the quadratic equation, you obtain:

L=48 inches

6. Finally, you must substitute the value of the length, into W=24+L:

W=24+L
W=24+48
W=72 inches

7. Therefore, the dimensions of the screen are:

L=48 inches
W=72 inches

4 0

2 years ago