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4 years ago

Find the value of the indicated angles. PLEASE HELP!!

Mathematics

1 answer:

SSSSS [86.1K]4 years ago

7 0

Answer:

Part 1) The measure of the angle is 47°

Part 2) The measure of the angle is 52°

Step-by-step explanation:

Part 1)

we know that

The inscribed angle is half that of the arc it comprises.

we have that

(12y-1)°=(9y+11)° ----> because the arc that the inscribed angles comprise is the same

Solve for y

12y-9y=11+1

3y=12

y=4°

Find the measure of the angle

(12y-1)°=12(4)-1=47°

Part 2)

we know that

The inscribed angle is half that of the arc it comprises.

we have that

2(3m+2)°=(4m+20)° ----> because the arc that the inscribed angles comprise is the same

Solve for m

6m+4=4m+20

6m-4m=20-4

2m=16

m=8

Find the measure of the angle

(4m+20)°=4(8)+20=52°

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Simplify the Expression I WILL GIVE BRAINLYIST!!!!!

sweet-ann [11.9K]

1+5×3/2
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Your answer is option A. 8.

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3 years ago

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Find the distance between the points (-7, -6) and (-3, 0).

FrozenT [24]

Answer:

7.21 unit

Step-by-step explanation:

distance= √[(-7+3)^2 + (-6-0)^2]

= 7.21 unit

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3 years ago

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

3 years ago

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A shirt is originally priced at $24.50 a shirt goes on sale for 30% off the original price how much money to the nearest cent do

Ainat [17]

Answer:

7.86 is the answer

Step-by-step explanation:

30% of 24.50 is 7.35.

7.35+ 7% is 7.86

7.86 to the nearest cent is 7.86

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3 years ago

Determine whether the statement is true or false. If f and g are positive increasing functions on an interval I, then fg is incr

Leno4ka [110]

Answer: false

Step-by-step explanation:

If f and g are increasing on I, this implies that f' > 0 on I and g' > 0 on I. That is both f' and g' have a positive slope. However,

Using product rule;

(fg)' = fd(g) + gd(f)

(fg)' = f * g' + f' * g

and although it is given that g' and f' are both positive we don't have any information about the sign of the values of the functions themselves(f and g). Therefore, if at least one of the functions has negative values there is the possibility that the derivative of the product will be negative. For example;

f = x, g = 5x on I = (-5, -2)

f' = 1 and g' =5 both greater than 0

f and g are both lines with positive slopes therefore they are increasing, but f * g = 5x^2 is decreasing on I.

6 0

3 years ago