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3 years ago

Which of the following statements are true about the polynomial?

Mathematics

1 answer:

Pani-rosa [81]3 years ago

3 0

Answer:

Step-by-step explanation:

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Julie has 1.54 L of water in a large bottle for a science experiment. She pours the water into beakers. Each beaker holds 0.22 L

RoseWind [281]

Answer:

7 beakers

Step-by-step explanation:

1.54/.22=7

6 0

3 years ago

-4x-2y=-12 4x+8y=-24

Alla [95]

Answer:

$x=6\\ \\y=-6$

Step-by-step explanation:

Given the system of two equations:

$\left\{\begin{array}{l}-4x-2y=-12\\ \\4x+8y=-24\end{array}\right.$

Add these two equations (left side add to left side and right side add to right side):

$-4x-2y+4x+8y=-12+(-24)\\ \\-2y+8y=-12-24\\ \\6y=-36\\ \\y=-6$

Substitute $y=-6$ into the first equation:

$-4x-2\cdot (-6)=-12\\ \\-4x+12=-12\\ \\-4x=-12-12\\ \\-4x=-24\\ \\x=6$

8 0

3 years ago

Henry is 3 times older than julia. rhianna is 1/2 as old as julia. what formula shows the relationship between henry and rhianna

Lubov Fominskaja [6]

H=henry's age
j=julia's age
r=rhianna's age

h is 3 times of j
h=3j

r is 1/2 times of j
r=(1/2)j or
2r=j

so

h=3j
2r=j
sub 2r for j
h=3(2r)
h=6r

henry's age is 6 times of rhianna's age

5 0

3 years ago

Read 2 more answers

Simplify 32 ⋅ 35. (4 points)<br> 37<br> 310<br> 97<br> 910

mafiozo [28]

Answer:

1120

Mutiply the numbers then add

5 0

3 years ago

Read 2 more answers

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago