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2 years ago

What is the vertex of this equation?

Mathematics

1 answer:

AfilCa [17]2 years ago

7 0

Answer:

(5, 4)

Step-by-step explanation:

The equation can be written in vertex form by adding and subtracting 25:

y = -(x^2 -10x +25) -21 +25

y = -(x -5)^2 +4

Then the vertex can be identified as (5, 4).

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The Tigers, a football team, must gain 10 yards in the next four plays to keep possession of the ball. The Tigers lose 24 yards,

Nikolay [14]

Answer:

<h3>B. The must gain 20 yards to keep possession</h3>

Step-by-step explanation:

Let the number of yards gained be positive

Let the number of yards lost be negative

Since the total that must be gained in the next four games is 10, hence the total is +10

If The Tigers lose 24 yards, gain 4 yards, and lose 10 yards in the first three rounds with x being the number of yards in the fourth round, this is expressed as;

-24 + 4 - 10 + x = 10

-20 - 10 + x = 10

-30 + x = -10

x = -10 + 30

x = 20

Hence they need 20 yards in the last round to keep possession

7 0

2 years ago

15) -1 - 2mx =6yx solve for x

tekilochka [14]

Answer : - 1 / 6y + 2m
The full solution is in the photo:)
( sorry if it’s wrong ;-; )

8 0

2 years ago

Solve 4 • (−6). (1 point) a −42 b −24 c 10 d 24

Furkat [3]

Answer:

its 400 c

Step-by-step explanation:

because it is

6 0

2 years ago

the cylinder has radius 4√3 and hight h. the total surface area of the cylinder is 56π√6. find the exact value of h giving the a

Arte-miy333 [17]

The area of the cylinder is a function of its height (h) and radius, ( $\mathbf{4\sqrt 3}$ )

The exact value of h is: $\mathbf{7\sqrt 2- 4\sqrt 3}$

The given parameters are:

$\mathbf{Area =56\pi\sqrt 6}$

$\mathbf{r=4\sqrt 3}$

The surface area of a cylinder is calculated as:

$\mathbf{Area = 2\pi rh + 2\pi r^2}$

Substitute values for Area

$\mathbf{56\pi\sqrt 6= 2\pi rh + 2\pi r^2}$

Divide through by pi

$\mathbf{56\sqrt 6= 2 rh + 2r^2}$

Substitute value for r

$\mathbf{56\sqrt 6= 2 (4\sqrt 3)h + 2(4\sqrt 3)^2}$

$\mathbf{56\sqrt 6= 8h\sqrt 3 + 2\times 48}$

$\mathbf{56\sqrt 6= 8h\sqrt 3 + 96}$

Collect like terms

$\mathbf{8h\sqrt 3 = 56\sqrt 6- 96}$

Make h the subject

$\mathbf{h = \frac{56\sqrt 6}{8\sqrt 3}- \frac{96}{8\sqrt 3}}$

$\mathbf{h = 7\sqrt 2- \frac{12}{\sqrt 3}}$

Rationalize

$\mathbf{h = 7\sqrt 2- \frac{12\sqrt 3}{3}}$

$\mathbf{h = 7\sqrt 2- 4\sqrt 3}$

Hence, the exact value of h is: $\mathbf{7\sqrt 2- 4\sqrt 3}$