Which problem can be represented by the equation 12z + 2= 98

Which of the following could be the ratio of the length of the longer leg of a 30-60-90 triangle to the length of its hypotenuse

Aliun [14]

Answer:

Option C is correct.

Ratio of longer leg to hypotenuse is; $\sqrt{3} : 2$

Step-by-step explanation:

This is the special right angle triangle 30°-60°-90° as shown below in the figure.

The side opposite the 30° angle is always the shortest because 30 degrees is the smallest angle.

The side opposite the 60° angle will be the longer leg, because 60 degrees is the mid-sized degree angle in this triangle.

Finally , the side opposite the 90° angle will always be the largest side(Hypotenuse) because 90 degrees is the largest angle.

In 30°−60°−90° right triangle,

the length of the hypotenuse is twice the length of the shorter leg,also
the length of the longer leg is $\sqrt{3}$ times the length of the shorter leg.

Then:

the sides are in proportion i.e, $1:\sqrt{3} :2$

Therefore, the ratio of the length of the longer leg to the length of its hypotenuse is: $\sqrt{3} : 2$

7 0

2 years ago

Read 2 more answers

Please help........................

xxTIMURxx [149]

My calculator gave me 22 but download Calculate84 and see if you get the same answers

8 0

2 years ago

What is the equation of the following line? Be sure to scroll down first to see all answer options. (0,3) (3,0) A. y = -x + 3 B.

qaws [65]

The correct answer is Choice A.

If you plot the points on a graph, you will see that there is a slope of -1 and the y-intercept is (0, 3).

This matches the equation of y = -x + 3 in Choice A.

4 0

2 years ago

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

2 years ago

a square painting has an area of 81x^2-90x-25. A second square painting has an area of 25x^2+30x+9. What is an expression that r

galina1969 [7]

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A1 ------> the area of the first square painting

A2 ----> the area of the second square painting

D -----> the difference of the areas

we have

$A1=81x^{2}-90x-25$

$A2=25x^{2}+30x+9$

case 1) The area of the second square painting is greater than the area of the first square painting

The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting

D=A2-A1

$D=(25x^{2}+30x+9)-(81x^{2}-90x-25)$

$D=(-56x^{2}+120x+34)$

case 2) The area of the first square painting is greater than the area of the second square painting

The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting

D=A1-A2

$D=(81x^{2}-90x-25)-(25x^{2}+30x+9)$

$D=(56x^{2}-120x-34)$

4 0

2 years ago