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2 years ago

I NEED UR HELPFNGKFNKFJFF

Mathematics

1 answer:

alisha [4.7K]2 years ago

8 0

For Q3, you can subsitute x = 2 and y = 5 into the systems

-5(2) + 5 = -5

so it's a solution for the first system

-4(2)+2(5) = 2

it's also a solution for the second system

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On a quiz worth 5 points, two students earned a 5, zero students earned a 4, zero students earned a 3, zero students earned a 2,

olchik [2.2K]

The class average would be all the test scores added up and divided by the number of scores.

5+5+1+1+1+1+1+0+0+0+0= 15

15/11=1.37

The class average is 1.37

8 0

4 years ago

Read 2 more answers

A cone has a diameter of 24 in. and a slant height of 15 in.

Fiesta28 [93]

<span>first of all that do you think the answer is</span>

6 0

3 years ago

Triangle ABC with vertices A(-2, -3), B(5, -1), and C(2, 2) is translated by (x,y) → (x - 1, y + 2). Then the image, triangle A'

Elan Coil [88]

Answer:

A"(-1, -2)
B"(6, 0)
C"(3, 3)
(x, y) ⇒ (x+1, y+1)

Step-by-step explanation:

Translation vectors add, and the addition is commutative and associative.

The first translation adds (-1, 2) to the original coordinates. The second translation adds (2, -1) to the original coordinates. The two translations together add ...

(-1, 2) +(2, -1) = (-1+2, 2-1) = (1, 1)

to the original coordinates.

The single rule representing this translation is ...

(x, y) ⇒ (x +1, y +1)

Then the doubly-translated coordinates are ...

A(-2, -3) ⇒ A"(-1, -2)

B(5, -1) ⇒ B"(6, 0)

C(2, 2) ⇒ C"(3, 3)

8 0

2 years ago

This is my question... please help!

laila [671]

Answer:

if sinx=3/5=P/H

using pythogoras thoeren find B

B²= H²-P²

B²= 25-9

B=4

Tanx = P/B= 3/4

n(S)= 6

n(O) = 3 {1,3,5)

P(O) = n(O)/n(s)= 3/6=1/2

Step-by-step explanation:

6 0

3 years ago

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

4 years ago