All categories

4 years ago

PLEASE HELP!! 30 POINTS

Mathematics

1 answer:

levacccp [35]4 years ago

3 0

Answer:

10 (D)

Step-by-step explanation:

do the Pythagorean theorem

You might be interested in

: 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

A line has a slope of -3 and includes the point (-5,4) and (-7,q). What is the value of q

serious [3.7K]

Answer:

q = 10

Step-by-step explanation:

$\text{slope} =-3=\frac{q-4}{-7-(-5)} =\frac{q-4}{-2} \\\Longrightarrow -3=\frac{q-4}{-2} \\\Longrightarrow \left( -3\right) \times \left( -2\right) =q-4\\\Longrightarrow 6=q-4\\\Longrightarrow q=6+4\\\Longrightarrow q=10$

8 0

2 years ago

What is the simplified form of the following expression? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0 2

finlep [7]

To solve such questions we need to know more about expression.

<h2 /><h2>Expression </h2>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations that are formed according to rules which are dependent on the context.

<h2 /><h2 /><h2>Given to us,</h2>

$2\sqrt[4]{16x}-2\sqrt[4]{2y}+3\sqrt[4]{81x}-4\sqrt[4]{32y}+5\sqrt[4]{x}-4\sqrt[4]{32y}+5\sqrt[4]{x}-16\sqrt[4]{2y}+13\sqrt[4]{4}-10\sqrt[4]{2y}+35\sqrt[4]{x}-18\sqrt[4]{2y}$

rearranging we get,

$=2\sqrt[4]{16x}+3\sqrt[4]{81x}+5\sqrt[4]{x}+5\sqrt[4]{x}+13\sqrt[4]{4}+35\sqrt[4]{x}-2\sqrt[4]{2y}-4\sqrt[4]{32y}-4\sqrt[4]{32y}-16\sqrt[4]{2y}-10\sqrt[4]{2y}-18\sqrt[4]{2y}$

we know that,

$\sqrt[4]{81}=3\\\sqrt[4]{16}=2\\\sqrt[4]{32}=2\sqrt[4]{2}\\$

therefore,

$=4\sqrt[4]{x}+9\sqrt[4]{x}+5\sqrt[4]{x}+5\sqrt[4]{x}+13\sqrt[4]{x}+35\sqrt[4]{x}-2\sqrt[4]{2y}-8\sqrt[4]{2y}-8\sqrt[4]{2y}-16\sqrt[4]{2y}-10\sqrt[4]{2y}-18\sqrt[4]{2y}$