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3 years ago

PLS ANSWER ASAP LOTS OF POINTS!!

Mathematics

1 answer:

laiz [17]3 years ago

3 0

<h2><u><em>BRAINLIEST PLZ ! YES? oki thank you :) </em></u><em> its 1, 2 and 5</em></h2>

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Solve for m. the square root of 3m+4=m

Nataly [62]

the answer is m = -2

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4 years ago

Can i pls get some help here with how to even solve it using clear steps?

kirill [66]

first off, let's split the triplet into two equations, then from there on we'll do substitution.

$\cfrac{y}{x-z}=\cfrac{x}{y}=\cfrac{x+y}{z}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{\cfrac{y}{x-z}=\cfrac{x}{y}\implies }y^2=\underline{x^2-xz} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{\cfrac{x}{y}=\cfrac{x+y}{z}\implies }xz=xy+y^2\implies \stackrel{\textit{substituting for }y^2}{xz=xy+(\underline{x^2-xz})}$

$2xz=xy+x^2\implies 2xz=x(y+x)\implies \cfrac{2xz}{x}=y+x \\\\\\ 2z=y+x\implies 2=\cfrac{y+x}{z}\implies 2=\cfrac{x+y}{z} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{}{ \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases}}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=2 \end{cases}\implies \begin{cases} \cfrac{y}{x-z}=2\\[2em] \cfrac{x}{y}=2 \end{cases}$

that of course, is only true if x + y, or our numerator doesn't turn into 0, if it does then our fraction becomes 0 and our equation goes south. Keeping in mind that x,y and z are numeric values that correlate like so.

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2 years ago

△ABC is mapped to △A′B′C′ using each of the given rules.

juin [17]

Answer:

The image is not loading!!

Step-by-step explanation:

Re-upload the question so maybe I can see it.

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3 years ago

Rotate each shape around the given axis and answer the questions below

JulsSmile [24]

Well if i had to make an educated guess my assumption would be similar to the hypothesis of this complicated situation now what i'm trying to say is i have no idea hope you find someone who can help you.

8 0

4 years ago

Find the area of a triangle whose vertices are located at (-2,5), (-4, -3), and (3,1). Evaluate the determinant using diagonals.

myrzilka [38]

Answer:

The area of the triangle is 48 unit²

Step-by-step explanation:

The given vertices of the triangle are;

(-2, 5), (-4, -3), and (3, 1)

The formula for finding the area of a triangle with given coordinates of the vertices is as follows;

$\Delta = \dfrac{1}{2}\times \begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} = \dfrac{1}{2}\times \left | x_1\cdot y_2 - x_2\cdot y_1 + x_2\cdot y_3 - x_3\cdot y_2 + x_3\cdot y_1 - x_1\cdot y_3\right |$

Substituting gives;

$\Delta = \dfrac{1}{2}\times \begin{vmatrix}-2& 5 & 1\\ -4 & -3 & 1\\ 3 & 1 & 1\end{vmatrix} \\\Delta = \dfrac{1}{2}\times \left | (-2)\times (-3) - ((-4)\times 5) + (-4)\times 1 - 3\times (-3) + 3 \times 5 - (-2)\times 1\right | \\\Delta = \dfrac{1}{2}\times \left | 6 +20 -4 +9 + 15+2\right | = 48$

The area of the triangle = 48 unit².

5 0

3 years ago