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miv72 [106K]
3 years ago
15

Can Someone Help Me With This?? :)

Mathematics
1 answer:
kykrilka [37]3 years ago
5 0
Rational) No
Integer) No
Whole) Yes
Natural) Yes
Describe the number:
It’s whole and Natural and it is 5
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find the value of each variable in the circle to the right. the dot represents the center of the circle.
gladu [14]

We have been given an image of a circle. We are asked to find the value of each variable.

We can see that angle b corresponds to diameter of circle. We know that the measure of an angle that is inscribed to the diameter of a circle is 90 degrees . Therefore, the value of b will be 90.

we can see that angle is an inscribed angle of arc 99 degrees. We know that measure of an inscribed angle is half the measure of inscribed arc.

a=\frac{1}{2}\cdot 99

a=49.5

Therefore, the value of a is 49.5 units.

We know that measure of all angles of a triangle is 180 degrees. The measure of 3rd angle will be half the measure of c, so we can set an equation as:

a+b+\frac{1}{2}c=180

49.5+90+\frac{1}{2}c=180

139.5+\frac{1}{2}c=180

139.5-139.5+\frac{1}{2}c=180-139.5

\frac{1}{2}c=40.5

\frac{1}{2}c\cdot 2=40.5\cdot 2

c=81

Therefore, the value of c is 81.

3 0
3 years ago
What is equivalent <br> to 2x-7(6)
trasher [3.6K]

-40 i think.............. im not positive but i did my best

6 0
3 years ago
-3x+18=7x what could you do to isolate the variable term to one side of the equation
Alexxandr [17]

Answer:

7x=−x+24 7 x = − x + 24  

Step-by-step explanation:

The equations we solved in the last section simplified nicely so that we could use the. Our strategy will involve choosing one side of the equation to be the variable side, and step by step, to isolate the variable terms on one side of the equation

7 0
3 years ago
Read 2 more answers
1680 at 12% for 6 months
vfiekz [6]

Answer:

Could you please expand on your question? Once you do I will edit the answer I put (this) to telling the answer. (The upcoming and this sentence is pre-written and copy and pasted in every brainly question or comment I write or answer.) If this answer helped you please consider giving it brainliest.

- •Trix•

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8 0
3 years ago
Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
Read 2 more answers
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