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3 years ago

Helpppppppppppppppppppppppppppppppppppp

Mathematics

1 answer:

yanalaym [24]3 years ago

6 0

Answer:

Step-by-step explanation:

a^3b^2c = b(a^3bc)

5bc = b(5c)

15a^2b = b(15a^2)

GCF = b

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Match the two step inequality to the correct graph.

miskamm [114]

Answer:

Hello,

I think -2p < - 10 goes with graph one

Then -M = 8 > 6 goes to Graph Two

Next 13 > 4r + 5 goes to Graph three

Last but not least -7 - Z < -1 goes to Graph Four

If I am Wrong i am sorry

Step-by-step explanation:

3 0

3 years ago

PLEASE HURRY IM BEING TIMED LOOK AT THE PICTURE

Andru [333]

Answer:

it is an enlargement with a scale factor greater than one.

Step-by-step explanation:

you can tell this, because the pre-image is smaller than the image, and its more than double the size :)

3 0

4 years ago

This is the problem Dan says 3/5 Is the same as 3.5. Is he correct? Explain

lesya692 [45]

Answer:

He is not correct

Explanation:

The first number is $\frac{3}{5}$

We can note that the numerator is less than the denominator. This means that the given fraction is less than 1

In decimal form, $\frac{3}{5}$ is equal to 0.6

The second number is 3.5

We can note that this number is greater than 1.

This means that the two numbers are not equal

Hope this helps :)

4 0

3 years ago

Read 2 more answers

A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after subm

True [87]

Answer:

A) cooling constant = 0.0101365

B) $\frac{df}{dt} = k ( 60 - F )$

c) F(t) = 60 + 77.46 $e^{0.0101365t}$

D)137.46 ⁰

Step-by-step explanation:

water temperature = 60⁰F

temperature of Bar after 20 seconds = 120⁰F

temperature of Bar after 60 seconds = 100⁰F

A) Determine the cooling constant K

The newton's law of cooling is given as

= $\frac{df}{dt} = k(60 - F)$

= ∫ $\frac{df}{dt}$ = ∫ k(60 - F)

= ∫ $\frac{df}{60 - F}$ = ∫ kdt

= In (60 -F) = -kt - c

60 - F = $e^{-kt-c}$

60 - F = $C_{1} e^{-kt}$ ( note : $e^{-c}$ is a constant )

after 20 seconds

$C_{1}e^{-k(20)}$ = 60 - 120 = -60

therefore $C_{1} = \frac{-60}{e^{-20k} }$ ------- equation 1

after 60 seconds

$C_{1} e^{-k(60)}$ = 60 - 100 = - 40

therefore $C_{1} = \frac{-40}{e^{-60k} }$ -------- equation 2

solve equation 1 and equation 2 simultaneously

= $\frac{-60}{e^{-20k} }$ = $\frac{-40}{e^{-60k} }$

= 6 $e^{20k}$ = 4 $e^{60k}$

= $\frac{6}{4} e^{40k}$ = In(6/4) = 40k

cooling constant (k) = In(6/4) / 40 = 0.40546 / 40 = 0.0101365

B) what is the differential equation satisfied

substituting the value of k into the newtons law of cooling)

60 - F = $C_{1} e^{0.0101365(t)}$

F(t) = 60 - $C_{1} e^{0.0101365(t)}$

The differential equation that the temperature F(t) of the bar

$\frac{df}{dt} = k ( 60 - F )$

C) The formula for F(t)

t = 20 , F = 120

F(t ) = 60 - $C_{1} e^{0.0101365(t)}$

120 = 60 - $C_{1} e^{0.0101365(t)}$

$C_{1} e^{0.0101365(20)}$ = 60

$C_{1} = 60 * 1.291$ = 77.46

C1 = - 77.46⁰ as the temperature is decreasing

The formula for f(t)

= F(t) = 60 + 77.46 $e^{0.0101365t}$

D) Temperature of the bar at the moment it is submerged

F(0) = 60 + 77.46 $e^{0.01013659(0)}$

F(0) = 60 + 77.46(1)

= 137.46⁰

5 0

4 years ago

Two-thirds of Tim’s class has cupcakes.One half of the cupcakes have chocolate frosting.what fraction of students have cupcakes

natulia [17]

The fraction of students having cupcakes with chocolate frosting is $\frac{1}{3} x$

Explanation:

Given:

Let the number of students in class be x

According to the question:

Students having cupcake = $\frac{2}{3} x$

Frosting cupcakes = $\frac{1}{2} X \frac{2}{3}x$

Fraction of students having cupcakes with chocolate frosting = $\frac{1}{2} X \frac{2}{3}x$

= $\frac{1}{3}x$

Thus, the fraction of students having cupcakes with chocolate frosting is $\frac{1}{3} x$

7 0

3 years ago