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3 years ago

The value of 2 · 5^2 is 100. True or False?

Mathematics

1 answer:

rosijanka [135]3 years ago

7 0

1. False
2. There isn’t a variable, I cannot answer.

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Each side of a square is increasing at a rate of 7 cm/s. At what rate is the area of the square increasing when the area of the

Vilka [71]

Answer:

Step-by-step explanation:

The rate if increase of side of square is 7cm/s

dL/dt=7

At what area is the area increasing

dA/dt=?

Areas of square is give to be 36cm²

Analysis

Let L be the length of each side of the square, a square has equal side.

Area of square is give as L²

Area=L²

L²=36

i.e L=√36

L=6cm

Therefore at t=0 L=6cm and A=36cm²

A=L²

dA/dL=2L

Given that, dL/dt=7

Multiply the two differential together

dA/dt= dA/dL × dL/dt

dA/dt= 2L × 7

dA/dt= 2×6×7

dA/dt= 84cm²/s

The rate of increase of the square is 84cm²/s

4 0

3 years ago

Which of the following trigonometric ratios has a value that is undefined?

masya89 [10]

Answer:

it is sin C

Step-by-step explanation:

3 0

4 years ago

A cricket team has a 0.25 chance of losing

Westkost [7]

Answer:

a) 0.125

b) 0.015635

c) 0.00000095367431640625‬

Step-by-step explanation:

a) $0.25^2 = 0.125$

b) $0.25^3 = 0.015625$

c) $0.25^{10} = 0.00000095367431640625\\\\$

8 0

3 years ago

Read 2 more answers

Carter drawers one side of equilateral triangle pqr on the coordinate plane at points P(-3, 2) and Q(5, 2). Which ordered pair i

Neko [114]

Given: Two coordinates of an equilateral triangle P(-3, 2) and Q(5, 2).

To find: Coordinate of third vertex R.

Solution: Let us take coordinate of R is (x,y).

Because we are given PQR is an equilateral triangle, all side of the triangle PQR are equal.

We can say PQ = QR = RS.

Let us find equations by using distance formula.

Distance between two coordinates is given by formula

$Distance = \sqrt{(x2-x1)^2+(y2-y1)^2}$

$PQ=\sqrt{(5-(-3))^2+(2-2)^2} = \sqrt{(5+3)^2+(0)^2} = \sqrt{8^2+0} = \sqrt{64} =8$

$QR=\sqrt{(x-5)^2+(y-2)^2} = \sqrt{x^2-10x+25 + y^2-4y+4} = \sqrt{x^2+y^2-10x-4y+29}$

$RS=\sqrt{(x-(-3)^2+(y-2)^2} = \sqrt{x^2+6x+9 + y^2-4y+4} = \sqrt{x^2+y^2+6x-4y+13}$ .

We know, PQ= QR and PQ= RS.

\sqrt{x^2+y^2-10x-4y+29}[/tex] = 8

Squaring both sides, we get

$x^2+y^2-10x-4y+29 =64$ -------- equation (1)

and \sqrt{x^2+y^2+6x-4y+13}[/tex] = 8

Squaring both sides, we get

$x^2+y^2+6x-4y+13=64$ ---------- equation (2).

Subtracting equation 2 from equation 1.

[/tex]x^2+y^2-10x-4y+29[/tex] = 64

[/tex]x^2+y^2+6x-4y+13[/tex] = 64

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

-16x +16 = 0

Subtracting 16 on both sides ,

-16x +16-16 = 0-16

-16x = -16

Dividing -16 on both sides.

-16x/-16 = -16/-16

x=1.

Plugging x=1 in first equation

$(1)^2+y^2-10(1)-4y+29 = 64$

$1+y^2-10-4y+29 = 64$ (Simplifying)

$y^2-4y+20 = 64$

Subtracting 64 from both sides.

$y^2-4y+20-64 = 64-64$

$y^2-4y-44 = 0$

By using quadratic formula..

$y=\frac{-(-4)+\sqrt{(-4)^2-4(1)(-44)} }{2(1)} and$

$y=\frac{-(-4)-\sqrt{(-4)^2-4(1)(-44)} }{2(1)} [tex]y=2\left(1+2\sqrt{3}\right),\:y=2\left(1-2\sqrt{3}\right)$

In decimal form

y=8.928, -4.928.

Therefore, coordinates of vertex R could be (1, 8.928) or (1, -4.928).

7 0

3 years ago

In need of help pls. Try and answer asap.

Alina [70]

Answer:

Step-by-step explanation:

∠1 and 85 are supplementary

m∠1 = 180 - 85 = 95°

4 0

3 years ago