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3 years ago

Which quadrilaterals have two pairs of opposite sides that are parallel and have no right angles?

Mathematics

2 answers:

prohojiy [21]3 years ago

8 0

Your answer would be C

Grace [21]3 years ago

5 0

A trapezoid does not have 2 pair of parallel sides so we can cross that out. A rhombus does not have right angles so cross that out. So we got A and C. A square has 2 pair of parallel lines and and 4 right angle. Your answers should be.
<span>C. Square
It could not be A. Most likely its C. So you can pick both or just C.</span>

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Heidi and Andrew designed two raised flowerbeds for their garden. Heidi's flowerbed was 5 feet long by 3 feet wide, and Andrew's

d1i1m1o1n [39]

Answer:

$V=90\ ft^3$

Step-by-step explanation:

-Given the two heights are similar:

#Heidi's flowered bed dimensions and volume is calculated as:

$v=lwh$

$=5\times 3\times 2\\\\=30\ ft^3$

#The width of Andrew's is twice Heidi's. The width and volume is thus calculated as:

$w_a=2\times 3=6 \ ft\\\\V_a=lwh\\\\=5\times 6 \times 2\\\\=60\ ft^3$

#Total soil volume of the soil used is determined by summing the two different flowerbed volumes:

$V=V_h+V_a\\\\=30+60\\\\=90 \ ft^3$

7 0

3 years ago

PLZ HELP WITH THIS QUESTION!

Bad White [126]

Answer:

Step-by-step explanation:

Given the quadratic

d = - 16t² + 12t ← subtract d from both sides

- 16t² + 12t - d = 0 ← in standard form

with a = - 16, b = 12, c = - d

Use the quadratic formula to solve for t

t = ( - 12 ± $\sqrt{12^2-(4(-16)(-d))}$ ) / - 32

= ( - 12 ± $\sqrt{144-64d}$ ) / - 32

= ( - 12 ± $\sqrt{16(9-4d)}$ ) / - 32

= ( - 12 ± 4 $\sqrt{9-4d}$ ) / - 32

= $\frac{-12}{-32}$ ± $\frac{4}{-32}$ $\sqrt{9-4d}$

= $\frac{3}{8}$ ± $\frac{1}{8}$ $\sqrt{9-4d}$

= $\frac{3}{8}$ ± $\frac{\sqrt{9-4d} }{8}$ → D

8 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

What expression can be used to add 5/12 + 3/8 PLZ TELL ME

Julli [10]

Answer:

19/24

Step-by-step explanation:

5/12 + 3/8

LCM: 24

(5×2 + 3×3)/24

(10 + 9)/24

19/24

5 0

2 years ago

Plz help I’m getting timed

Fed [463]

id say its a rhombus. all the other shapes can have right angles

6 0

2 years ago

Read 2 more answers