If you start with a principal amount of $150 and the simple interest rate is 3%. What will the

Please help with these 2 answers i only have 10 minutes left

natta225 [31]

Answer:

1/4 of a pound
4/15 of the flat

Step-by-step explanation:

For number 1:

3/4 ÷ 3 = 0.25
0.25 = 1/4

For number 2:

4/5 ÷ 3 = 4/15

I hope this helps!

5 0

2 years ago

Read 2 more answers

Trisha and Byron are washing and vacuuming cars to raise money for a class trip. Trisha raised $38 washington 5 cars and vacuumi

ANEK [815]

Answer:

They charged $6 for washing and $2 for vacuuming the car.

Step-by-step explanation:

Let the charge of washing a car be 'x'.

And also let the charge of vacuuming a car be 'y'.

Now according to question, Trisha raised $38 washing 5 cars and vacuuming 4 cars.

So framing the above sentence in equation form, we get;

$5x+4y=38\ \ \ \ \ equation\ 1$

Again according to question, Byron raises $28 by washing 4 cars and vacuuming 2 cars.

So framing the above sentence in equation form, we get;

$4x+2y=28\ \ \ \ \ equation\2$

Now multiplying equation 2 by '2', we get;

$2(4x+2y)=28\times2\\\\8x+4y=56\ \ \ \ equation\ 3$

Now we subtract equation 1 from equation 3 and get;

$(8x+4y)-(5x+4y)=56-38\\\\8x+4y-5x-4y=18\\\\3x=18\\\\x=\frac{18}{3}=6$

Now substituting the value of 'x' in equation 1, we get;

$5x+4y=38\\\\5\times6+4y=38\\\\30+4y=38\\\\4y=38-30=8\\\\y=\frac{8}{4}=2$

Hence They charged $6 for washing and $2 for vacuuming the car.

8 0

3 years ago

How are rate of change and slope related for a linear relationship?

ale4655 [162]

Hello! In slope-intercept form, y=mx+b. M is the slope and b is the y-intercept. The rate of change and slope are interchangeable, therefore, they both create a line through the slope-intercept form. Feel free to rate my answer Brainliest if I helped you!

7 0

3 years ago

Select the correct answer,

krok68 [10]

There is no picture

4 0

2 years ago

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0

3 years ago