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3 years ago

6

determine the equation of the straight line which passes through the point 1 and 5 and is perpendicular to the line 3x + y is eq

ual to 4

1 answer:

xenn [34]3 years ago

7 0

The straight line is perpendicular to y = -3x + 4.

Therefore, the gradient of the straight line must be 3.

The equation of the straight line is y = 3x + 2.

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What solid fits snugly through each shape when held in the right position -square, triangle, circle

Nana76 [90]

Answer:

Im pretty sure its circle

7 0

3 years ago

Read 2 more answers

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$

6 0

2 years ago

In a right triangle one leg measures a quarter of the other, how much do its acute angles measure? with procedure please

KATRIN_1 [288]

Answer:

The answer to your question is 14° and 76°

Step-by-step explanation:

Data

leg 1 = x

leg 2 = x/4

angle 1 = ?

angle 2 = ?

Process

To solve problem use trigonometric functions. We must use the trigonometric function that relates both legs (tangent).

tangent Ф = Opposite side / Adjacent side

-Substitution

tan Ф = (x/4) / x

-Simplification

tan Ф = x / 4x

tan Ф = 1/4

- Find Ф

Ф = tan⁻¹(1/4) = 14°

- Find the other acute angle

The sum of the internal angles in a triangles equals 180°

Ф + Ф₁ + 90° = 180

-Solve for Ф₁

Ф₁ = 180 - 90 - 14

-Result

Ф₁ = 76°

3 0

3 years ago

Please solve both with explanation, I’ll give brainliest

DiKsa [7]

1st and 2nd pictures are for finding area and 3rd picture is for finding perimeter

hope you understood well and if not and you can ask again

4 0

3 years ago

A basketball court is 94 feet by 50 feet. Eight fans ca

IgorC [24]

Answer:

The entire floor can contain approximately 4178 fans

Step-by-step explanation:

The first step is to calculate the area of the basketball floor.

we can do this by multiplying the length by the breadth as such 94 X 50 = 4700 square feet.

The second step is to calculate the area occupied by 8 of the fans. We can do this by multiplying 3 ft by 3ft = 9 square feet.

From this, it will be easier to estimate the area occupied by only one fan. This can be got by dividing 9 square feet by 8.

This is = 1.125 square feet.

To get the number of students it can occupy, we divide the total area of the court by the area occupied by one student.

4700/ 1.125 =4177.8 $\approx$ 4178 fans

4 0

3 years ago