All categories

2 years ago

Discounted 40%. A weed eater originally sells for $145. How much will the weed eater

Mathematics

1 answer:

mote1985 [20]2 years ago

4 0

100% = 145
-40%
60%= 145 x 0.6
It costs $87

You might be interested in

PLEASE HURRY IM TIMED!!!!!

notka56 [123]

Answer:

0.4 kmffffffffffffffffffffffffffffffffff

4 0

3 years ago

Read 2 more answers

What two numbers add to -5 and multiply to 12?

Deffense [45]

Negative 6 and negative 2

3 0

3 years ago

Read 2 more answers

An oil rig has a height of 74.98 m visible above water and extends 518.16 m below the surface to the sea floor. What is the tota

pashok25 [27]

Answer:

So you know that the height that is visible above x=0 is 74.98 m
From x=0 below it has a total height of 518.16m to the sea floor

That tells us that we have two parts account for

74.98+518.16m = 593.14 m

8 0

3 years ago

Read 2 more answers

The figure below shows a line graph and two shaded triangles that are similar:

ella [17]

Answer:

Step-by-step explanation:

h wh I h6j2tbwtbt2byyh3ht36

7 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D

Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago