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Softa [21]
3 years ago
14

Five times 9/12 - 1/4 + 0.5

Mathematics
1 answer:
zalisa [80]3 years ago
6 0

Answer:

The answer is 4

Step-by-step explanation:

u welcome boo

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1) Given: circle k(O), ED= diameter ,m∠OEF=32°, m(arc)EF=(2x+10)° Find: x
qaws [65]

1. The major arc ED has measure 180 degrees since ED is a diameter of the circle. The measure of arc EF is (2x+10)^\circ, so the measure of arc DF is

m\widehat{DF}=360^\circ-180^\circ-(2x+10)^\circ=(170-2x)^\circ

The inscribed angle theorem tells us that the central angle subtended by arc DF, \angle DOF, has a measure of twice the measure of the inscribed angle DEF (which is the same angle OEF) so

m\angle DOF=2m\angle OEF=64^\circ

so the measure of arc DF is also 64 degrees. So we have

170-2x=64\implies106=2x\implies\boxed{x=53}

###

2. Arc FE and angle EOF have the same measure, 56 degrees. By the inscribed angle theorem,

m\angle EOF=2m\angle EDF\implies56^\circ=2m\angle EDF\implies m\angle EDF=28^\circ

Triangle DEF is isosceles because FD and ED have the same length, so angles EFD and DEF are congruent. Also, the sum of the interior angles of any triangle is 180 degrees. It follows that

m\angle EFD+m\angle EDF+m\angle DEF=180^\circ\implies\boxed{m\angle EFD=76^\circ}

Triangle OFE is also isosceles, so angles EFO and FEO are congruent. So we have

m\angle EFO+m\angle FEO+m\angle EOF=180^\circ\implies\boxed{m\angle EFO=62^\circ}

7 0
3 years ago
A = √7 + √c and b = √63 + √d where c and d are positive integers.
mrs_skeptik [129]

Answer:

\displaystyle a:b=\frac{1}{3}

Step-by-step explanation:

<u>Ratios </u>

We are given the following relations:

a=\sqrt{7}+\sqrt{c}\qquad \qquad[1]

b=\sqrt{63}+\sqrt{d}\qquad \qquad[2]

\displaystyle \frac{c}{d}=\frac{1}{9} \qquad \qquad [3]

From [3]:

9c=d

Replacing into [2]:

b=\sqrt{63}+\sqrt{9c}

We can express 63=9*7:

b=\sqrt{9*7}+\sqrt{9c}

Taking the square root of 9:

b=3\sqrt{7}+3\sqrt{c}

Factoring:

b=3(\sqrt{7}+\sqrt{c})

Find the ration a:b:

\displaystyle a:b=\frac{\sqrt{7}+\sqrt{c}}{3(\sqrt{7}+\sqrt{c})}

Simplifying:

\boxed{a:b=\frac{1}{3}}

3 0
3 years ago
Find the value of the rectangular prism
Ainat [17]
27.

Volume = length*width*height
So 3*3*3=27
6 0
3 years ago
I need help with this iready question
Ad libitum [116K]

Answer:

64%

Step-by-step explanation:

The percent that was plastic is the plastic over the total * 100 %

percent plastic = 16/25 *100%

                          = .64 *100%

                            64%

7 0
3 years ago
Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter
Vaselesa [24]

We have been given a function f(x)=2x^3+3x^2-336x. We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

f'(x)=2(3)x^{2}+3(2)x^1-336

f'(x)=6x^{2}+6x-336

6x^{2}+6x-336=0

x^{2}+x-56=0

x^{2}+8x-7x-56=0

(x+8)-7(x+8)=0

(x+8)(x-7)=0

x=-8,x=7

So x=-8,7 are critical points and these will divide our function in 3 intervals (-\infty,-8)U(-8,7)U(7,\infty).

Now we will find derivative over each interval as:

f'(x)=(x+8)(x-7)

f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16

Since f'(9)>0, therefore, function is increasing on interval (-\infty,-8).

f'(x)=(x+8)(x-7)

f'(1)=(1+8)(1-7)=(9)(-6)=-54

Since f'(1), therefore, function is decreasing on interval (-8,7).

Let us check for the derivative at x=8.

f'(x)=(x+8)(x-7)

f'(8)=(8+8)(8-7)=(16)(1)=16

Since f'(8)>0, therefore, function is increasing on interval (7,\infty).

(b) Since x=-8,7 are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

f(-8)=2(-8)^3+3(-8)^2-336(-8)

f(-8)=1856

f(7)=2(7)^3+3(7)^2-336(7)

f(7)=-1519

Therefore, (-8,1856) is a local maximum and (7,-1519) is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

f''(x)=6(2)x^{1}+6

f''(x)=12x+6

12x+6=0

12x=-6

\frac{12x}{12}=-\frac{6}{12}

x=-\frac{1}{2}

Therefore, x=-\frac{1}{2} is an inflection point of given function.

3 0
3 years ago
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