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3 years ago

Someone please help me with this

Mathematics

2 answers:

zalisa [80]3 years ago

5 0

Answer:

B: up

Step-by-step explanation:

The +b in this function is considered the y-intercept. When you Increase this value, you are adding that much to the function. This means that the value will go up.

34kurt3 years ago

5 0

The answer is B.

Hope this helps!

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Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds a

kirill115 [55]

Answer:

12 feet per second.

Step-by-step explanation:

Please consider the complete question.

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be represented $h(t)=-16t^2+20t$ .

What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?

We will use average rate of change formula to solve our given problem.

$\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}$

$\text{Average rate of change}=\frac{h(\frac{1}{2})-h(0)}{\frac{1}{2}-0}$

$\text{Average rate of change}=\frac{-16\cdot(\frac{1}{2})^2+20\cdot \frac{1}{2}-(-16\cdot(0)^2+20\cdot (0))}{\frac{1}{2}-0}$

$\text{Average rate of change}=\frac{-16\cdot\frac{1}{4}+10-(0)}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{-4+10}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{6}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{2\cdot 6}{1}$

$\text{Average rate of change}=12$

Therefore, Spot's average rate of ascent is 12 feet per second.

3 0

3 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: