All categories

3 years ago

Plz help 20 points and more

Mathematics

2 answers:

tino4ka555 [31]3 years ago

8 0

I believe the answer is c and have a very nice day

Lilit [14]3 years ago

3 0

The answer is C
By multiplying -65 and 8, you get -520
And I'm pretty sure there is no need to eliminate in this equation

You might be interested in

The graph of f(x) = x + 1 is shown in the figure. Find the largest δ such that if 0 < |x – 2| < δ then |f(x) – 3| < 0.4

MArishka [77]

SO what you need to do is:
Start with |f(x) - 3| < 0.4
and plug in f(x) = x+1
to get |f(x) – 3| < 0.4
|x+1 – 3| < 0.4
|x - 2| < 0.4
-0.4 < x - 2 < 0.4
-0.4+2 < x < 0.4+2
1.6 < x < 2.4
So delta would be 2.3
Hope this is what you were looking for

8 0

3 years ago

Read 2 more answers

What is a ratio can someone give me more than one answer please for this question

Ymorist [56]

The number relation that is between two amounts.

7 0

3 years ago

If 5(p + 3) = 18 + 2p, what is the value of p?

Rina8888 [55]

Answer:

p=1

Step-by-step explanation:

5(p+3)=18+2p

5p+15=18+2p

5p-2p=18-15

3p=3

p=3

p=1

5 0

3 years ago

Read 2 more answers

Answer Part A and Part B for the triangle shown

Tom [10]

Answer:

for Part A

Answer is C.

Step-by-step explanation:

$\tan(a) = \frac{opposite}{adjacent } \\$

Here the angle, a is 60°, Opposite is 20 and adjacent is x.

Therefore we have,

$\tan(60 ) = \frac{20}{x}$

Making x the subject gives us

$x = \frac{20}{ \tan(60) }$

Part B

Value of x to the nearest tenth

$x = 11.547 \\ to \: the \: nearest \: tenth \\ x = 12$

5 0

2 years ago

Find the derivative of |x|/(x-1)

ki77a [65]

$\bf \cfrac{|x|}{x-1}\iff \cfrac{\sqrt{x^2}}{x-1}\iff \cfrac{(x^2)^{\frac{1}{2}}}{x-1}
\\\\\\
\textit{using the quotient rule}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x(x-1)-(x^2)^{\frac{1}{2}}\cdot 1}{(x-1)^2}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}}}{(x-1)^2}\implies \cfrac{dy}{dx}=\cfrac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}(x-1)^2}
\\\\\\\cfrac{dy}{dx}=\cfrac{x(x-1)-|x|}{|x|(x-1)^2}$

4 0

3 years ago