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2 years ago

Help me with my math homework

Mathematics

1 answer:

saul85 [17]2 years ago

6 0

1. 2√2
2. √5
3. 5√3
4. 10
5. 4√2
6. 3√5
7. 4√3
8. 5√5
9. 3√2
10. 3√3
11. 4
12. 6
13. 5√2
14. 2√3

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4. Gabby makes banana bread at her bakery every Monday and Wednesday. Her

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Answer:

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The ratio is 5 (sugar) : 50 (butter)...

You can make the unit rate: 1 cup of sugar: 10 tbs of butter

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3 years ago

Can someone help me? In this net, the two triangles are right triangles. All quadrilaterals are rectangles. What is its surface

Nikitich [7]

Answer:

Net Area = 168

Step-by-step explanation:

See attachment for complete question

I'll make reference to the attachment, when needed.

The shapes that make up the net are labelled 1 to 5.

So, to calculate the surface area of the net, we simply calculate the surface area of each label.

Label 1: Rectangle

Area = Length * Width

Area = 10 * 5

Area = 50

Label 2: Rectangle

Area = Length * Width

Area = 8 * 5

Area = 40

Label 3: Rectangle

Area = Length * Width

Area = 6 * 5

Area = 30

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Area = ½Base * Height

Base = 6 and Height = 8

So:

Area = ½ * 6 * 8

Area = 24

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Area = ½Base * Height

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So:

Area = ½ * 6 * 8

Area = 24

So, the net area is the summation of the calculated areas of label 1 to 5

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2 years ago

Which parent functions go through the origin? Select all that apply.

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2 years ago

Read 2 more answers

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

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3 years ago