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2 years ago

Please find the slope

Mathematics

1 answer:

olga nikolaevna [1]2 years ago

7 0

After using the slope formula your answer would be 5/0.1 but when simplifying your answer is 50

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Area and percent .

siniylev [52]

Answer:

a. 61.92 in²

b. 21.396 ≈ 21.4%

c. $4.71

Step-by-step explanation:

a. Amount of waste = area of rectangular piece of stock - area of two identical circles cut out

Area of rectangular piece of stock = 24 in × 12 in = 288 in²

Area of the two circles = 2(πr²)

Use 3.14 as π

radius = ½*12 = 6

Area of two circles = 2(3.14*6²) = 226.08 in²

Amount of waste = 288 - 226.08 = 61.92 in²

b. % of the original stock wasted = amount of waste ÷ original stock × 100

= 61.92/288 × 100 = 6,162/288 = 21.396 ≈ 21.4%

c. 288 in² of the piece of stock costs $12.00,

Each cut-out circle of 113.04 in² (226.08/2) will cost = (12*113.04)/288

= 1,356.48/288 = $4.71.

3 0

3 years ago

Anyone have any idea how to do this???

bagirrra123 [75]

Answer:

Step-by-step explanation:

3 0

2 years ago

Solve the two-step equation 14=31.7-3x

ch4aika [34]

Answer: X = 5.9

Step-by-step explanation:

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

Twenty-seven is smaller than five times a number increased by seven

mixas84 [53]

Answer:

27(less than symbol) 5x+7

Step-by-step explanation:

5 times a number is 5x

increase means add

4 0

2 years ago