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2 years ago

Can somebody please help me?!! I will give BRAINLIEST!!

Mathematics

1 answer:

Wewaii [24]2 years ago

5 0

Answer:

Step-by-step explanation:

Mark the two points (-1,7) and (1,-1) on the graph. Then draw a straight line between them. To determine the equation that goes through these two points, we can use the two given points to find the slope of the line. The standard form of a straight line equation is

y = mx + b,

where m is the slope and y is the y-intercept (the value of y when x = 0).

Slope is also known as the "Rise"/"Run" - the change in y divided by the change in x. We can use the two points to calculate this:

Rise (-1-(7) = -8 Run = (1 - (-1) = 2

The slope is therefore (-8/2) or -4.

y = -4x + b

We can find b by entering either of the two points in y = -4x + b and solve for b. I'll use (1,-1) since I have my 1's multiplication table memorized

y = -4x + b

-1 = -4(1) + b

b = 3

The straight line equation that connects the two points is

y = -4x + 3

You can graph this equation (e.g., on DESMOS) to see how it intersects the points. <u>[Attached]</u>

The coordinates of the y intercept are (0,3).

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Factorize 1-(3x-1)^2

solniwko [45]

Step-by-step explanation:

(1+3x-1)×{1-(3x-1)}

=3x.(1-3x+1)

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6 0

3 years ago

I need help finding the area

aniked [119]

<h3>Given :</h3>

Base of triangle = 7 yd
Height of triangle = 10 yd

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Area of triangle

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We know:-

When base and height of triangle is given we use this formula:

$\bigstar \boxed{ \rm Area \: of \: triangle = \frac{base \times height}{2} }$

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So:-

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{base \times height}{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 10}{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times2 }{2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times\cancel2 }{\cancel2} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times1 }{1} \\$

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$\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5$

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$\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2} \\$

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$\therefore \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} { \red{\bf{yd} }^{ \red2} }}$

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$\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$ ]