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2 years ago

Given: ΔPSQ, PS = SQ Perimeter of ΔPSQ = 50 SQ – PQ = 1 Find: Area of ΔPSQ

Mathematics

1 answer:

Lana71 [14]2 years ago

8 0

Answer:

120 units²

Step-by-step explanation:

Perimeter = PS + SQ + PQ

50 = SQ + SQ + (SQ -1)

51 = 3SQ

17 = SQ

17 -1 = 16 = PQ

The midpoint of the base is one leg of the right triangle whose other leg is the height of this isosceles triangle. That height is ...

h = √(17² -(16/2)²) = √225 = 15

Then the area is ...

A = (1/2)bh = (1/2)(16)(15) = 120 . . . . . square units

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{3x=7y=5 {5x-2y=-19

Anton [14]

Answer:

(-3, 2)

Step-by-step explanation:

{3x=7y=5

{5x-2y=-19

solve....

maybe it was 3x + 7y = 5

and 5x - 2y = -19

2*(3x + 7y = 5)

7*(5x - 2y = -19)

-------------

6x + 14y = 10

35x -14y = -133

41x = -123

x = -3

y = ...

5*(-3) - 2y = -19

-15 - 2y = -19

2y = 4

y = 2

(-3, 2)

7 0

2 years ago

What is the answer to m diveded by 8-7=12

lesya [120]

We simplify the equation to the form, which is simple to understand
m/8-7=12

Simplifying:
 + 0.125m-7=12

We move all terms containing m to the left and all other terms to the right.
 + 0.125m=+12+7

We simplify left and right side of the equation.
 + 0.125m=+19

We divide both sides of the equation by 0.125 to get m.
m=152

Hope this helps :)♥

7 0

3 years ago

Read 2 more answers

mr. Smith is putting in a new floor in his bathroom he has already filled 25 ft.² of floor measuring 10‘ x 6‘ what percent has h

Salsk061 [2.6K]

Answer:

41.67%

Step-by-step explanation:

$tiled \: percent = \frac{25}{10 \times 6} \times 100 \\ \\ = \frac{2500}{60} \\ \\ = 41.6666667\% \\ \\ \approx \: 41.67\% \\$

7 0

2 years ago

Which of the following is a quadratic?

irina1246 [14]

Number 3 i would say i think it is

6 0

2 years ago

Given the position of the particle, what the position(s) of the particle when it’s at rest

choli [55]

The position function of a particle is given by:

$X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t$

The velocity function is the derivative of the position:

$\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}$

The particle will be at rest when the velocity is 0, thus we solve the equation:

$2t^2-9t-18=0$

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

$t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Substituting:

$\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}$

We have two possible answers:

$\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}$

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

$undefined$

6 0

1 year ago