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4 years ago

I don’t get this at all can someone help?

Mathematics

1 answer:

aleksklad [387]4 years ago

5 0

Reasons (In order):

Given
Given
Definition of a midpoint
Reflexive property
SAS
CPCTC

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The equation y = 3x + 16 can be used to find the salary, y for a person with x years experience. How many years experience would

fgiga [73]

Y=106,000. 106,000=3x+16. 106,016=3x now divide.... 106,016÷3=x. _____ 35338.666667=x

5 0

3 years ago

Please help<br><br> Check answers

Law Incorporation [45]

Answer:

Number 8 the answer is 120

Number 10 answer is 115

Step-by-step explanation:

8. 180-60=120

it's because the straight line is 180 then you just need to solve it use this

10. 180-65=115

the reason same like number 8

Hope this answer can help you to solve the math problems

6 0

3 years ago

The circumference of a circle is 23π m. What is the area, in square meters? Express your answer in terms of \piπ

NARA [144]

Answer:

Step-by-step explanation:

The circumference of a circle is 23π

On the other hand, The circumference of a circle = 2×π×r ; where r is the radius.

then

by equating the last two expressions we get: 23π = 2πr then r = 23/2

then

the area = π × r² = (23/2)²×π = (132.25)×π

7 0

3 years ago

Read 2 more answers

F(x)=x^2-3x+18 how many distinct real number zeros does f have?

vampirchik [111]

Answer:

None of the distinct real number zeros the function have.

Step-by-step explanation:

Considering the function

$f(x)=x^2-3x+18$

The zeroes of a function are those values that touches the x-axis. In order to find those values we must

Set $f(x) = 0$ in order to find those values

$0=x^2-3x+18$

$\mathrm{Switch\:sides}$

$x^2-3x+18=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-3,\:c=18:\quad x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}=3+\sqrt{63}i$

$x=\frac{3+\sqrt{63}i}{2\cdot \:1}$

$x=\frac{3+3\sqrt{7}i}{2}$

$x=\frac{3}{2}+\frac{3\sqrt{7}}{2}i$

Similarly,

$x=\frac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}:\quad \frac{3}{2}-i\frac{3\sqrt{7}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{3}{2}+i\frac{3\sqrt{7}}{2},\:x=\frac{3}{2}-i\frac{3\sqrt{7}}{2}$

BUT, NONE OF THEM ARE REAL NUMBER ZEROS.

Therefore, none of the distinct real number zeros the function have.

5 0

4 years ago

PROVE THAT:<br><br>cos 20° - sin 20° = \sqrt{2}sin25°<br><br>

Lemur [1.5K]

Answer:

See below.

Step-by-step explanation:

$\cos(20)-\sin(20)=\sqrt{2}\sin(25)$

First, use the co-function identity:

$\sin(90-x)=\cos(x)$

We can turn the second term into cosine:

$\sin(20)=\sin(90-70)=\cos(70)$

Substitute:

$\cos(20)-\cos(70)=\sqrt{2}\sin(25)$

Now, use the sum to product formulas. We will use the following:

$\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$

Substitute:

$\cos(20)-\cos(70)=-2\sin(\frac{20+70}{2})\sin(\frac{20-70}{2})\\\cos(20)-\cos(70) =-2\sin(45)\sin(-25)\\\cos(20)-\cos(70)=-2(\frac{\sqrt{2}}{2})\sin(-25)\\ \cos(20)-\cos(70)=-\sqrt{2}\sin(-25)$

Use the even-odd identity:

$\sin(-x)=-\sin(x)$

Therefore:

$\cos(20)-\cos(70)=-\sqrt{2}\sin(-25)\\\cos(20)-\cos(70)=-\sqrt{2}\cdot-\sin(25)\\\cos(20)-\cos(70)=\sqrt{2}\sin(25)$

Replace the second term with the original term:

$\cos(20)-\sin(20)=\sqrt{2}\sin(25)$

Proof complete.

4 0

4 years ago