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3 years ago

Can you help me please (get brainliest)

Mathematics

1 answer:

marta [7]3 years ago

6 0

9514 1404 393

Answer:

4/25

Step-by-step explanation:

Put the numbers where the letters are and do the arithmetic.

$\left(2\cdot\dfrac{-1}{2}\cdot\dfrac{2}{5}\right)^2=\left(\dfrac{-2}{5}\right)^2=\dfrac{(-2)^2}{5^2}=\boxed{\dfrac{4}{25}}$

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Jayden and Sheridan both tried to find the missing side of the right triangle. A right triangle is shown. One leg is labeled as

stellarik [79]

Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

$a=7\ cm\\c=13\ cm$

substitute

$13^{2}=7^{2}+b^{2}$

Solve for b

$169=49+b^{2}$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

we have that

Jayden's Work

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\b=13\ cm$

substitute and solve for c

$7^{2}+13^{2}=c^{2}$

$49+169=c^{2}$

$218=c^{2}$

$c=\sqrt{218}\ cm$

$c=14.76\ cm$

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

Sheridan's Work

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\c=13\ cm$

substitute

$7^{2}+b^{2}=13^{2}$

Solve for b

$49+b^{2}=169$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

therefore

Sheridan's Work is correct

6 0

3 years ago

Read 2 more answers

Find the slope of (1, 2) (-3, 5) SHOW ME STEP BY STEP

Sloan [31]

-3/4

Step-by-step explanation:

f(x) = (1,2) & (-3,5)

f(x) = y²-(y¹)

_____

x²-(x¹)

y=5-2

___

-3-1

y=-3/4

7 0

2 years ago

Read 2 more answers

: Find the approximate perimeter of the isosceles triangle A OPD.

AleksAgata [21]

Answer:

24.98 units

Step-by-step explanation:

The picture of the question in the attached figure

we have the coordinates

P(1,-6),D(4,3),O(7,-6)

The perimeter of triangle OPD is equal to

$P=OP+PD+OD$

the formula to calculate the distance between two points is equal to

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

step 1

Find the distance OP

we have

O(7,-6),P(1,-6)

substitute in the formula

$d=\sqrt{(-6+6)^{2}+(1-7)^{2}}$

$d=\sqrt{(0)^{2}+(-6)^{2}}$

$d_O_P=6\ units$

step 2

Find the distance PD

we have

P(1,-6),D(4,3)

substitute in the formula

$d=\sqrt{(3+6)^{2}+(4-1)^{2}}$

$d=\sqrt{(9)^{2}+(3)^{2}}$

$d_P_D=\sqrt{90}\ units$

step 3

Find the distance OD

we have

O(7,-6),D(4,3)

substitute in the formula

$d=\sqrt{(3+6)^{2}+(4-7)^{2}}$

$d=\sqrt{(9)^{2}+(-3)^{2}}$

$d_O_D=\sqrt{90}\ units$

step 4

Find the perimeter

$P=6+\sqrt{90}+\sqrt{90}$

$P=6+9.49+9.49=24.98\ units$

4 0

3 years ago

Hello, can someone help me with this problem?

exis [7]

Answer:

Area of Rectangle A

$Area = 4x^2$

Area of Rectangle B

$Area = 2x^2$

Fraction

$Fraction =\frac{2}{3}$

Step-by-step explanation:

From the attached, we understand that:

The dimension of rectangle A is 2x by 2x

The dimension of rectangle B is x by 2x

Area of rectangle is calculated as thus;

$Area = Length * Breadth$

Area of Rectangle A

$Area = 2x * 2x$

$Area = 4x^2$

Area of Rectangle B

$Area = x * 2x$

$Area = 2x^2$

Area of Big Rectangle

The largest rectangle is formed by merging the two rectangles together;

The dimension are 3x by 2x

The Area is as follows

$Area = 2x * 3x$

$Area = 6x^2$

The fraction of rectangle A in relation to the largest rectangle is calculated by dividing area of rectangle A by area of the largest rectangle;

$Fraction = \frac{Rectangle\ A}{Biggest}$

$Fraction =\frac{4x^2}{6x^2}$

Simplify

$Fraction =\frac{2x^2 * 2}{2x^2 * 3}$

$Fraction =\frac{2}{3}$

7 0

3 years ago

Someone help wh ores.

kirill115 [55]

Answer

Point 1 (2, 8)

Point 2 (5, 20)

Step-by-step explanation:

Since we are given the x values of 2 and 5, substitute them in the equation and you will get 8 and 20.

3 0

2 years ago