All categories

3 years ago

WILL GIVE BRAINLIEST ! 12 POINTS !

Mathematics

2 answers:

galben [10]3 years ago

4 0

1 over 2, 3 over 2 is the correct answer.

fenix001 [56]3 years ago

4 0

The answer would be (5/2),(7/2). Sorry it took a while to get to your answer!

You might be interested in

HELP!! Two mechanics worked on a car. The first mechanic worked for 20 hours, and the second mechanic worked for 5 hours. Togeth

joja [24]

Answer:

first mechanic: $105/hour
second mechanic: $55/hour

Step-by-step explanation:

Let r represent the rate charged by the first mechanic. Then 160-r is the rate charged by the second mechanic. The total of charges is ...

20r +5(160-r) = 2375

15r +800 = 2375 . . . . . . eliminate parentheses

15r = 1575 . . . . . . . . . . . . subtract 800

r = 105 . . . . . . . . . . . . . . . divide by 15. First mechanic's rate.

160-r = 55 . . . . . Second mechanic's rate

The first mechanic charged $105 per hour; the second, $55 per hour.

7 0

3 years ago

Read 2 more answers

Which of the following numbers is written in scientific notation?

nata0808 [166]

Answer:

i think is b

Step-by-step explanation:

8 0

2 years ago

Hi i just wanna know if i'm right on this math question<br><br> Thank you.

Stolb23 [73]

Answer:

Correct my guy or gal.

Step-by-step explanation:

6 0

3 years ago

What is the solution to the system of equations? x=4y 6 x-3y=4

ahrayia [7]

If you graph the two equations then you will find that they intersect at (-2,-2).

You then have to check your answer.

x=4y+6
(-2)=4(-2)+6
-2=-8+6
-2=-2

x-3y=4
(-2)-3(-2)=4
-2+6=4
4=4

8 0

3 years ago

Find the approximate area between the curve f(x) = -4x² + 32x and on the x-axis on the interval [0,8] using 4 rectangles. Use th

Doss [256]

Split up the interval [0, 8] into 4 equally spaced subintervals:

[0, 2], [2, 4], [4, 6], [6, 8]

Take the right endpoints, which form the arithmetic sequence

$r_i=2+\dfrac{8-0}4(i-1)=2i$

where 1 ≤ <em>i</em> ≤ 4.

Find the values of the function at these endpoints:

$f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i$

The area is given approximately by the Riemann sum,

$\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i$

where $\Delta x_i=\frac{8-0}4=2$ ; so the area is approximately

$\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}$

where we use the formulas,

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

6 0

3 years ago