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3 years ago

(1.)

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

6 0

1. x= -13 x=13
2. (-5,-3)
3. x=-6 x=6
4. x=-5/6 x=5/6

Anna35 [415]3 years ago

4 0

Gotcha!

The 1st answer is x=-13 or x=13

The 2nd answer is -5<y<-3

The 3rd answer is t=-6 or t=6

The 4th answer is a=-1/8 or a=1/8

You might be interested in

) Find the coordinates of the point A' which is the symmetric point

soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is $A'(\dfrac{31}{5}, \dfrac{18}{5})$ .

7 0

2 years ago

Going top to bottom label them as A, B, C, and D

Ket [755]

I hope this helps you

7 0

3 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. N(10) 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. N(10) 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

Solve the equation for 0° ≤ x ≤ 90°. Round to the nearest degree. tan^2 B = 4

Anettt [7]

ANSWER

$B=63\degree$

EXPLANATION

We want to solve the trigonometric equation;

$tan^2(B)=4$ , where $0\degree \le B \le 90\degree$ .

Taking square root of both sides gives,

$tan(B)=\pm \sqrt{4}$

$tan(B)=\pm 2$

Since B is in the first quadrant, we proceed with;

$tan(B)=2$ , because the tangent ratio is positive in the first quadrant.

$\Rightarrow B =arctan(2)$

$\Rightarrow B =63.43\degree$ .

Correct to the nearest degree,

$B =63\degree$

7 0

3 years ago

Write an expression in expression in simplest form that represents the total amount in the situation. You rent x pairs of shoes

lesya692 [45]

2x+(1.50+9)= $10.50×2= 21 so x=21

8 0

3 years ago