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eimsori [14]
3 years ago
6

5 tens + 6 ones = 4 tens + ___ ones

Mathematics
1 answer:
charle [14.2K]3 years ago
5 0

Answer:

16 ones

Step-by-step explanation:

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I don't understand how to do this. Can anyone help. Preferably with steps. Thank you in advance!
Andrew [12]
You put the x=6
6³-4(6)²-12(6)+15=15
3 0
3 years ago
PLEASE HELP ITS DUE SOON IM SO CONFUSED! HELP NEEDED!!
ddd [48]

Answer:

<u>e</u>

Step-by-step explanation:

According to Angle Sum Property :

<u>∠A + ∠B + ∠C = 180°</u>

<u />

============================================================

Given :

⇒ ∠A = 50°

⇒ ∠B = 32°

===========================================================

Solving :

⇒ 50° + 32° + ∠C = 180°

⇒ ∠C + 82° = 180°

⇒ ∠C = <u>98°</u>

6 0
2 years ago
What is the result if 3x-9 is evaluated when x = -5?​
Anton [14]

Answer:

Step-by-step explanation:

I wonder if it is the minus that is causing the problem?

Let x = - 5

3*(-5) - 9

-15 - 9

This is a money question. If you are 15 dollars in debt and you spend another 9, where are you? (In the company of the American government. They do this all the time).

- 15 - 9 = - 24

5 0
3 years ago
Let u = &lt;-9, 4&gt;, v = &lt;8, -5&gt;. Find u - v. <br> URGENT
Pie

To subtract vectors, you simply have to subtract correspondent coordinates.

So, if you have

u = \langle u_1,\ u_2,\ldots u_n\rangle ,\quad v = \langle v_1,\ v_2,\ldots v_n\rangle

the subtraction is simply

u-v = \langle u_1-v_1,\ u_2-v_2,\ldots u_n-v_n\rangle

So, in your case, we have

\langle-9,4\rangle-\langle 8,-5\rangle=\langle -9-8, 4+5\rangle=\langle -17,9\rangle

3 0
3 years ago
Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.
guapka [62]

\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}

\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill

7 0
2 years ago
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