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bogdanovich [222]

3 years ago

6

Paul wrote the equation t=2m+40 to represent the temperature, t, in degrees Celsius, after a substance had been heated for m min

utes.

1 answer:

Usimov [2.4K]3 years ago

4 0

First you need to fill in the variables with what you already know, so
t=2m+40.
You don’t have enough information to solve this problem. You need to have only 1 variable to solve it or you need more information

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What is the surface area of a rectangular prim 3mm,9mm 13mm

padilas [110]

SA = 2<span>(<span><span><span>wl</span>+<span>hl</span></span>+<span>hw</span></span><span>)

SA = 2((3)(13) + (9)(13) + (9)(3))

Multiply:

SA = 2(39 + 117 + 27)

Add:

SA = 2(183)

Multiply:

SA = 366</span></span>

4 0

3 years ago

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Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$\bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2$

i.e, 53 + 53 = 106

Hence, the triangle is a right - angled triangle as well.

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4 years ago

A line passes through the point (-4, -2) and has a slope of -5/2. Write an equation in slope- intercept form for this line.

kirill115 [55]

Answer:

<h2> y = -⁵/₂x - 12 </h2>

Step-by-step explanation:

The point-slope form of the equation is y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope:

m = -⁵/₂

(-4, -2) ⇒ x₀ = -4, y₀ = -2

The point-slope form of the equation:

y + 2 = -⁵/₂(x + 4)

So:

y + 2 = -⁵/₂x - 10 {subtract 2 from both sides}

y = -⁵/₂x - 12 ← the slope-intercept form of the equation

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