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3 years ago

Which would be a correct first step to solve the following system of linear equations using the elimination method? 2x+3y=6

Mathematics

1 answer:

Mrac [35]3 years ago

7 0

Answer:

Subtract

Step-by-step explanation:

Subtract to eliminate the y value so all that is left to slove is the x value. Once you have the x value, you can plug it in to find the y value.

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30 + 6p = 7 (p+ 6 ) -5 Please answer

melamori03 [73]

Answer: p=-7

Step-by-step explanation:

To solve for p, you want to isolate the variable, get p alone. You would use different algebraic properties to do so.

30+6p=7p+42-5

30+6p=7p+37

-7=p

p=-7

6 0

3 years ago

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A triangle has side lengths of 10 cm, 24 cm, and 32 cm. Classify the triangle as acute, obtuse, right of not a triangle.

Stolb23 [73]

It would be obtuse.

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3 years ago

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What is (-2)(-5)? Dhdjfhfhhdhd

lbvjy [14]

Answer:

$( - 2)( - 5) \\ - 2 \times - 5 \\ = - 10$

<h2>please mark me brainliest</h2>

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3 years ago

Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30

Dafna11 [192]

Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

$\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}$

$\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------$

$\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o$

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3 years ago

Finding the function of y=-2/3x +2

nordsb [41]

Answer:2 x + 3 y = 6

Step-by-step explanation:

in standard form

7 0

3 years ago