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3 years ago

PLS HELP HELP ME WITH THE QUESTION

Mathematics

1 answer:

Hunter-Best [27]3 years ago

8 0

Answer:

hi there!

the correct answer to this question is: D. reflection over the y-axis

Step-by-step explanation:

point a remains constant while the other 2 turned negative

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Function 1: y = 4x + 5

vlabodo [156]

Y=mx+b
m=slope

given
y=4x+5
m=4=slope

2.
for the points (x1,y1) and (x2,y2), the slope of the line passing trhough those points is (y2-y1)/(x2-x1)
given
(1,6) and (3,10)
slope=(10-6)/(3-1)=4/2=2

function 1 slope is 4
function 2 slope is 2
4>2

the first one has greater rate of change

6 0

3 years ago

Read 2 more answers

A quarter is 1/5 of an inch thick and 5 quarters weigh one ounce. Stephanie built a toothpick bridge that could safely hold 3 po

AleksandrR [38]

Answer:

Maximum height of quarter tower = 9.6 inches.

Step-by-step explanation:

The thickness of a quarter = $\frac{1}{5}$ inches

Weight of a quarter = 1 ounce

The maximum weight that could safely hold = 3 pounds.

<u>The conversion factor for ounce to pound is 1 ounce = 0.0625 pound</u>

The number of quarter used in the tower = no of ounces equivalent to 3 pounds.

No of ounces equivalent to 3 pounds = $\frac{3}{0.0625}$

No of ounces equivalent to 3 pounds = 48

So the height of quarter tower = 48 \times $\frac{1}{5}$ inches

= 9.6 inches.

4 0

3 years ago

What is the total amount of sap the trees produced that day ?

Vikki [24]

Answer:

1 gallon

Step-by-step explanation:

hope this helps

6 0

3 years ago

What is the zero of the linear function graphed below?

In-s [12.5K]

Answer:

-2

hope this helps you so much

6 0

2 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

5 0

3 years ago