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3 years ago

HELP

Mathematics

1 answer:

Aleks [24]3 years ago

3 0

Answer:

c) 6

Step-by-step explanation:

This is a straight-forward application of the Law of Cosines:

... a² = b² + c² -2bc·cos(A)

... a² = 8² +11² -2·8·11·cos(32.2°) ≈ 64 +121 -176·0.8462

... a² ≈ 36.07000

... a ≈ 6.00583

The best choice is c) 6.

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The variable z varies jointly with x and y and inversely with v and w. Using the constant of variation,k, from the last problem,

antiseptic1488 [7]

Answer:

Step-by-step explanation:

5 0

3 years ago

What is -√12+3√3? I need to know.

Gelneren [198K]

Answer:

√3

Step-by-step explanation:

√12=2√3

-2√3+3√3 = √3

Use calculator to estimate if necessary: 1.7320508075688773

4 0

3 years ago

Read 2 more answers

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

Urgent need of assistance

ikadub [295]

Answer:i got 329

Step-by-step explanation:

4 0

2 years ago

Can someone help me figure out the answer?

olga_2 [115]

Answer:

y = -2x +15

Step-by-step explanation:

The point-slope form of the equation for a line through (h, k) with slope m is ...

... y - k = m(x - h)

For your point (h, k) = (5, 5) and slope m = -2, the equation in point-slope form is ...

... y - 5 = -2(x - 5)

Simplifying, we get

... y - 5 = -2x +10

Adding 5 puts the equation into slope-intercept form, as you want.

... y = -2x +15

8 0

3 years ago