I need help with b and c please and thank you!!

How can you use mental math to find the percent of a number?

olga_2 [115]

Every percent of that number is 1/100 of that number
Example 30, 1/100 of 30 is .30
Example 100, 1/100 of 100 is 1
Example 52, 1/100 of 52 is .52

3 0

3 years ago

Which table represents a linear function?

Sladkaya [172]

Answer:

C

Step-by-step explanation:

C is the only function that have a consistent decrease while A is a trigonometric function, B is a non linear function, D is an exponential function

8 0

4 years ago

Recall the equation for a circle with center ( h , k ) and radius r . At what point in the first quadrant does the line with equ

astraxan [27]

Answer:

The Point in the first quadrant is $(\sqrt[2]{2.0268} ,6.559)$ .

Step-by-step explanation:

Since

we know the general equation of the circle whose center is at (h,k) and radius is r that is

$(x-h)^2+(y-k)^2 =r^2$

we have given center is (0,3) and radius is r= 4

substituting in the general equation of the circle we get

$x^2 +(y-3)^2 = 16$

Now substituting the y= 2.5x + 3 in the above equation we get

$x^2 +(2.5x + 3-3)^2 = 4^2$

$x^2 +6.25x^2 = 16$

here subtracting 16 from both sides and also perform addition

we get

$7.25x^2 = 16$

dividing both sides by 7.25 we get

$x^2 =2.2068$

taking square root both sides we get

$x= \frac{+}{-}\sqrt[2]{2.0268}$

but we need positive value of x because both x and y are positive in the first quadrant

that is

$x= \sqrt[2]{2.0268}$

substituting this into the given equation of the line we get

y = 2.5( $\sqrt[2]{2.0268}$ )+3

y = 6.559

so the point in the first quadrant is ( $\sqrt[2]{2.0268}$ ,6.559)

that is required

6 0

4 years ago

Read 2 more answers

Fifteen more than half a number is 9

Sidana [21]

15 + n = 9 (n equals "number")

8 0

4 years ago

Match the function with its graph.

kondor19780726 [428]

Answer:

Option a. 1C, 2A, 3B, 4D.

Step-by-step explanation:

1) We know that tan(x)=sin(x)/cos(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=0. For that reason, we know that the graph passes through the point (0,0).

If x=45, then sin(45)= $\frac{\sqrt{2}}{2}$ and cos(45)= $\frac{\sqrt{2}}{2}$ . Thus tan(45)=1. The only graph that passes through the point (0,0) and is possitive when x=45 is the graph C.

2) We know that cot(x)=cos(x)/sin(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=+∞. For that reason, we know that the graph has an asymptote in y=0, in other words, it never crosses the y-axis.

If x=45, then sin(45)= $\frac{\sqrt{2}}{2}$ and cos(45)= $\frac{\sqrt{2}}{2}$ . Thus cot(45)=1. The only graph that has an asymptote in y=0 and is possitive when x=45 is the graph A.

3) We know that -tan(x)=-sin(x)/cos(x). If x=0, sin(x)=0 and cos(x)=1 then -tan(x)=0. For that reason, we know that the graph passes through the point (0,0).

If x=45, then sin(45)= $\frac{\sqrt{2}}{2}$ and cos(45)= $\frac{\sqrt{2}}{2}$ . Thus -tan(45)=-1. The only graph that passes through the point (0,0) and is negative when x=45 is the graph B.

) We know that -cot(x)=-cos(x)/sin(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=-∞. For that reason, we know that the graph has an asymptote in y=0, in other words, it never crosses the y-axis.

If x=45, then sin(45)= $\frac{\sqrt{2}}{2}$ and cos(45)= $\frac{\sqrt{2}}{2}$ . Thus -cot(45)=-1. The only graph that has an asymptote in y=0 and is negative when x=45 is the graph D.

8 0

4 years ago