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soldi70 [24.7K]
3 years ago
13

Can someone help me please

Mathematics
2 answers:
Misha Larkins [42]3 years ago
6 0
I’m gonna try and help just hang on
stepladder [879]3 years ago
4 0
Y= x0.02

See how 1x.02= 0.2, 2x.02= .04. It’s multiplying the x times .02
You might be interested in
Factorise 6x^2 + 11x + 4​
zysi [14]
(3x+4)(2x+1)

If the question asks you to find roots/solutions:

3x + 4 = 0
3x = -4
x = -1.33333333

2x + 1 = 0
2x = -1
x = -0.5
4 0
2 years ago
Read 2 more answers
Can u help solve this ​
8_murik_8 [283]

Answer and Step-by-step explanation:

The expression is equal to 90.

This is because the lines that cut the right angle are cutting it in half, so those base angles are 45, leaving the center angle to be 90 (90 + 45 + 45 = 180).

90 = 19x + 4.5

85.5 = 19x

x = 4.5

<u>The value of x that makes the parallelogram a square is 4.5.</u>

<u></u>

<u></u>

<em><u>I hope this helps!</u></em>

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

6 0
2 years ago
GUYS PLS HELP MEEEE I REALLY NEED HELP
PilotLPTM [1.2K]

Answer:

x = 22.45

Step-by-step explanation:

Have a good day :)

6 0
3 years ago
the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2
lilavasa [31]

Answer:  c) \dfrac{3}{2} .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}

Given function : f(x) = x^2

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3

\Rightarrow\ f'(c)=3\ \ \ ...(i)

Since f'(x)=2x

then, at x=c, f'(c)=2c\ \ \ ...(ii)

From (i) and (ii), we have

2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}

Hence, the correct option is c) \dfrac{3}{2} .

4 0
3 years ago
Altitudes are congruent segments true or false
Snezhnost [94]
Not true, at least not always. When we are talking about an isosceles triangle there will be two equal altitudes and one is different. In case of a regular triangle all the other three altitudes have equal length! :)
6 0
3 years ago
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