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3 years ago

What is the approximate distance between point A (-8, 7) and point B (-2, 3)?

Mathematics

1 answer:

RoseWind [281]3 years ago

3 0

Answer:

7.2units

Step-by-step explanation:

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Do repeated addition

Step-by-step explanation:

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3 years ago

One number is seven more than four times another number. If the sum of the numbers is 22, find the numbers.

Agata [3.3K]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Area triangles? it says show work so pls do

coldgirl [10]

$\large{\underline{\underline{\pmb{\sf {\color {blue}{Solution:}}}}}}$

Given,

Area = 225 yd²

Base = 30 yd

Height = [To be calculated]

To find:

The height of the given triangle.

We know that, area of a triangle is:

$\frac{1}{2} \times (b \times h) \\ \\ \leadsto \frac{1}{2} \times (30 \times h) \\ \\ \leadsto \frac{1}{2} \times 30h \\ \\ \leadsto \frac{1}{ \cancel{2}} \times \cancel{30}h \\ \\ \leadsto1 \times 15h \\ \\ \leadsto \: h = 15 \: yd$

Therefore, the require height is 15 yd.

$\green\star$ Proof:

$\frac{1}{2} \times (b \times h) \\ \\ \leadsto \frac{1}{2} \times (30 \times 15) \\ \\ \leadsto \frac{1}{ \cancel{2}} \times \cancel{450} \\ \\ \leadsto1 \times 225 \\ \\ area = 225 {yd}^{2}$

$\boxed{ \frak \red{brainlysamurai}}$

6 0

2 years ago

Determine the effective annual yield for each investment.

Lyrx [107]

I think it’s B but don’t at me

3 0

3 years ago

PLEASE HELP I NEED THIS ASAP<br> Find the vertex of y=x^2-7x+4

Goshia [24]

Answer:

$\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

Step-by-step explanation:

$y=x^2-7x+4\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=1,\:b=-7,\:c=4\\x_v=-\frac{b}{2a}\\x_v=-\frac{\left(-7\right)}{2\cdot \:1}\\\mathrm{Simplify}\\x_v=\frac{7}{2}\\\mathrm{Plug\:in}\:\:x_v=\frac{7}{2}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4\\$

$\mathrm{Simplify\:}\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4:\quad -\frac{33}{4}\\y_v=-\frac{33}{4}\\Therefore\:the\:parabola\:vertex\:is\\\left(\frac{7}{2},\:-\frac{33}{4}\right)\\\mathrm{If}\:a0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=1\\\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

6 0

3 years ago