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2 years ago

Can anyone help me with this ?

Mathematics

1 answer:

Bad White [126]2 years ago

5 0

Answer:

b i think

Step-by-step explanation:

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3×-18=42 Solve for x!

Keith_Richards [23]

3x - 18 = 42
3x = 42 + 18
3x = 60
x = 60/3
x = 20

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3 years ago

36 + 120 : 12-63 What's this answer?

NeTakaya

Answer:

Step-by-step explanation:

36 + 120: 12-63

156 : 51

GCF of both sides are three. Divide both sides by three.

52:17

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2 years ago

What is 4.33 as a fraction

Scorpion4ik [409]

Answer: 4 1/3 or 13/3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If SDE ~ SWT, find WT.

faust18 [17]

Formula used: 40+x = 4x-3 x is used for ET

Step-by-step explanation:

40+x = 4x-3

+3 +3

43+x = 4x

-x -x

43 = 3x

3 3

14.3 = x

then substitude this on 5x+3 on WT and you arr good

5 0

2 years ago

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