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3 years ago

13

A soccer field is a rectangle 90 meters wide and 120 meters long. the coach asks the player to run from one corner to the corner

diagonally across. what is the distance?

1 answer:

melisa1 [442]3 years ago

7 0

A² + b² = c²
90² + 120² = c²
c² = 8100 + 14400
c² = 22500
c = √22500
c = 150

Answer: The diagonal distance is 150m long

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3 years ago

Can somebody help me with this problem

Mekhanik [1.2K]

Break the shape into 2 along the perforated line. now you have two rectangular prisms. find the volume of each and add them together to get your answer.

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3 years ago

The axis of symmetry for the graph of the function is f(x) = x2 + bx + 10 is x = 6. What is the value of b?

joja [24]

Answer:

$b=-12$

Step-by-step explanation:

we have

$f(x)=x^{2}+bx+10$

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex of the parabola

and the axis of symmetry is equal to $x=h$

In this problem we have the axis of symmetry $x=6$

so

the x-coordinate of the vertex is equal to $6$

therefore

For $x=6+1=7$ -----> one unit to the right of the vertex

Find the value of $f(7)$

$f(7)=7^{2}+b(7)+10$

$f(7)=59+7b$

For $x=6-1=5$ -----> one unit to the left of the vertex

Find the value of $f(5)$

$f(5)=5^{2}+b(5)+10$

$f(5)=35+5b$

Remember that

$f(5)=f(7)$ ------> the x-coordinates are at the same distance from the axis of symmetry

so

$59+7b=35+5b$ ------> solve for b

$7b-5b=35-59$

$2b=-24$

$b=-12$

3 0

3 years ago

Read 2 more answers

(sin teta + sec teta)^ + (cos teta+ cosec teta )^ = (1 + sec x cosec)^

ArbitrLikvidat [17]

$(sin\theta+sec\theta)^2+(cos\theta+cosec\theta)^2=(1+sec\theta\ cosec\theta)^2\\\\L=sin^2\theta+2sin\theta\cdot\frac{1}{cos\thewta}+\frac{1}{cos^2\theta}+cos^2\theta+2cos\theta\cdot\frac{1}{sin\theta}+\frac{1}{sin^2\theta}\\\\=(sin^2\theta+cos^2\theta)+\frac{2sin\theta}{cos\theta}+\frac{2cos\theta}{sin\theta}+\frac{1}{cos^2\theta}+\frac{1}{sin^2\theta}$

$=1+\frac{2sin^2\theta+2cos^2\theta}{sin\theta\ cos\theta}+\frac{sin^2\theta+cos^2\theta}{sin^2\theta\ cos^2\theta}\\\\=1+\frac{2(sin^2\theta+cos^2\theta)}{sin\theta\ cos\theta}+\frac{1}{sin^2\theta\ cos^2\theta}\\\\=1+\frac{2}{sin\theta\ cos\theta}+\frac{1}{sin^2\theta\ cos^2\theta}\\\\=1^2+2sec\theta\ cosec\theta+sec^2\theta\ cosec^2\theta\\\\=(1+sec\theta\ cosec\theta)^2=R$

4 0

3 years ago

Olenka [21]

Answer:

d = -1/3, 0

Step-by-step explanation:

Subtract the constant on the left, take the square root, and solve from there.

(6d +1)^2 + 12 = 13 . . . . given

(6d +1)^2 = 1 . . . . . . . . . .subtract 12

6d +1 = ±√1 . . . . . . . . . . take the square root

6d = -1 ±1 . . . . . . . . . . . .subtract 1

d = (-1 ±1)/6 . . . . . . . . . . divide by 6

d = -1/3, 0

_____

Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:

f(x) = (6x+1)^ +12 -13

We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.

4 0

3 years ago