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2 years ago

Moussa is preheating his oven before using it to bake. The initial temperature of the

Mathematics

1 answer:

Yuki888 [10]2 years ago

8 0

Answer:

365°, (65 +20t)°

Step-by-step explanation:

Temperature of oven

= initial temperature + 20°(number of minutes passed after it was turned on)

Temperature of oven after 15 minutes

= 65 +20(15)

= 65 +300

= 365°

Temperature after t minutes

= (65 +20t)°

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What is the answer to this question

astra-53 [7]

Answer: I think the answer is B

Step-by-step explanation:

8 0

3 years ago

I need help on this one asap pliz

Doss [256]

$C(5,4)(.40)^4(1-.40)^1=5(.4)^4(.6)^1=0.0768$

That's 7.68% = 7.7% (rounded)

Note: C(5, 4) is the number of combinations of 5 things taken 4 at a time. You may have seen this written as $_5C_4$

In general, $C(n,r)=\frac{n!}{r!(n-r)!}$

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3 years ago

Which of the following options have the same value as

Keith_Richards [23]

Question:

Which of the following options have the same value as 65% of 20?

Choose 2 answers:

(A) 0.65⋅20

(B) 65/100 divided by 20

(D) 65 * 20

Answer:

Option A has the same value as 65% of 20

Step-by-step explanation:

Let x be the value of 65% of 20

$x = 65\% of 20$

$x = \frac{65}{100} \times 20$

$x =0.65 \times 20$

$x =13$

Thus 65% of 20 is 13

Now ,

Solving Option A

=> $(0.65) \cdot (20)$

=> 13

Solving Option B

=> 65/100 divided by 20

=> $\frac{\frac{65}{100}}{20}$

=> $\frac{0.65}{20}$

=>0.0325

Solving Option C

=>65/20 * 100

=> $\frac{65}{20} \times 100$

=> $3.25 \times 100$

=>325

Solving Option D

=> 65 * 20

=> $65 \times 20$

=> 1300

8 0

3 years ago

Read 2 more answers

A rectangular piece of poster board measures 60 centimeters by 80 centimeters. Linn draws the net of a box on the poster board a

adell [148]

I am pretty sure it is 50 by 64 centimeters but if it is not I am really sory for the mistake.

3 0

3 years ago

Read 2 more answers

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

Alternative method

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

3 years ago