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joja [24]
3 years ago
14

Are all trapeziums rhombus. pls answer​

Mathematics
1 answer:
maksim [4K]3 years ago
8 0
No their not they have four sides but their not
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When a scatterplot is created from a table of values, which statement is correct
lorasvet [3.4K]
<span>It is possible for two points to have the same x-coordinate and the same y-coordinate. </span>
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3 years ago
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Can anyone help me with the rest of my ixl m.1 please and thank you
mina [271]

Answer:

u=54

Step-by-step explanation:

VX is middle line of triangle, so

VX=ZY/2.....(1)

We know that

Zy=u

Vx=u-27

Put that in (1) we have:

u-27=u/2

2u-54=u

2u-u=54

u=54

5 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I don't understand how to do this. :'(
natulia [17]

Answer:

Step-by-step explanation:

First and foremost, all quadratics have a domain of all real numbers (as long as we are not given only a portion of the graph, or one with endpoints. Our graph does not have endpoints, so it is assumed that the tails will continue to go down into negative infinity and at the same time, the x coordinates will keep growing as well.) Since our quadratic is upside down, it has a max. That means that none of the values on the graph will be above that point. All the values will be below that highest point (the highest y-value). Y-values indicate range, and since our highest y-value is at y = 2, then the range is

y ≤ 2

3 0
2 years ago
Find the area and the circumference of the circle. Round your answers to the nearest hundredth.
Alik [6]

Answer:

area = 200.96  \: {units}^{2}  \\ circumfrence = 50.24 \: units

Step-by-step explanation:

The formula to find the area and circumference of a circle is:

area = \pi {r}^{2}  \\ circumfrence = 2\pi \: r

<em>*</em><em>no</em><em>te that pi is 3.14*</em><em> </em>Therefore the area is:

formula = \pi {r}^{2}  \\ 3.14 \times  {8}^{2}  =  \\ 3.14 \times 64 = 200.96

Therefore the circumference is:

formula = \: 2\pi \: r \\ (2)(3.14)(8) =  \\ 3.14 \times 16 = 50.24

3 0
2 years ago
How do you do this question?
daser333 [38]

Answer:

B. 1/2

Step-by-step explanation:

\lim_{z \to 0} \frac{g(z)e^{-z}-3}{z^{2}-2z}

If we plug in 0 for z, we get 0/0.  Apply l'Hopital's rule.

\lim_{z \to 0} \frac{-g(z)e^{-z}+g'(z)e^{-z}}{2z-2}

Now when we plug in 0 for z, we get:

\frac{-g(0)e^{0}+g'(0)e^{0}}{2(0)-2}\\\frac{-g(0)+g'(0)}{-2}\\\frac{-3+2}{-2}\\\frac{1}{2}

4 0
2 years ago
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