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Elden [556K]
3 years ago
11

Identify the graph of the system of inequalities. 3 y? 2x n 10x + 20y < 300

Mathematics
2 answers:
boyakko [2]3 years ago
7 0

yesmfffStep-by-step explanation:f56t53433

6566666666

natali 33 [55]3 years ago
7 0

Answer: it’s the 1st one

Step-by-step explanation:

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Answer: 24

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Solve the exponential question . Leave your answer as a fraction
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problem → 2^x = 4, solve for x

⇒2^x=4

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Trevor covered a cylindrical can with paper for a project. The
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3 years ago
Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?
PIT_PIT [208]
A second degree polynomial function has the general form:

                  \displaystyle{f(x)=ax^2+bx+c, where a\neq0.

The leading coefficient is a, so we have a=-1.

5 is a double root means that :

i) f(5)=0,
ii) the discriminant D is 0, where D=b^2-4ac.

Substituting x=5, we have

                                        f(5)=a(5)^2+b(5)+c,

and since f(5)=0, and a is -1 we have:

                                          0=-25+5b+c
thus c=25-5b.


By ii) \displaystyle{b^2-4ac=0.

Substituting a with -1 and c with 25-5b we have:
                                     
          \displaystyle{b^2-4ac=0
          \displaystyle{b^2-4(-1)(25-5b)=0
          \displaystyle{b^2+4(25-5b)=0
          \displaystyle{b^2-20b+100=0
          \displaystyle{(b-10)^2=0
          \displaystyle{b=10 


Finally we find c: c=25-5b=25-50=-25

Thus the function is        \displaystyle{f(x)=-x^2+10x-25


Remark: It is also possible to solve the problem by considering the form

f(x)=-1(x-5)^2 directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is f(x)=a(x-r)^2
5 0
3 years ago
Read 2 more answers
20.) A=49.23 degrees, c=54.8Solve the right triangle. Express angles in decimal degrees.
vivado [14]

\begin{gathered} a=41.50 \\ b=35.78 \\ B=40.76\text{ \degree} \\  \end{gathered}

Explanation

Step 1

a) let

\begin{gathered} A=49.23\text{ \degree} \\ c=54.8\text{ \degree} \end{gathered}

b) b value

to find the measure of side b we can use cosine function

\begin{gathered} cos\theta=\frac{adjacent\text{ side}}{hypotenuse} \\ replace \\ cos\text{ 49.23=}\frac{b}{54.8} \\ b=54.8*cos49.23 \\ b=35.78 \end{gathered}

c) angle B

to find the measure of Angle B we can use sine function

\begin{gathered} sin\theta=\frac{opposite\text{  side}}{hypotenuse} \\ replace \\ sin\text{ B=}\frac{35.78}{54.8} \\ sin\text{ B= 0.65}\Rightarrow inverse\text{ function to isolate B} \\ B=\sin^{-1}(0.65) \\ B=40.76 \end{gathered}

d) side a

\begin{gathered} sin\theta=\frac{opposite\text{ side}}{hypotenuse} \\ sin\text{ A=}\frac{a}{c}=\frac{\placeholder{⬚}}{\placeholder{⬚}} \\ sin\text{ 49.23=}\frac{a}{54.8} \\ multiply\text{ both sides by 54.8} \\ 54.8s\imaginaryI n\text{49.23=}\frac{a}{54.8}*54.8 \\ 41.50=a \end{gathered}

so, the answer is

\begin{gathered} a=41.50 \\ b=35.78 \\ B=40.76\text{ \degree} \\  \end{gathered}

I hope this helps you

7 0
1 year ago
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