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Otrada [13]
3 years ago
13

Draw a shape with 4 sides and corners that is not a rectangle. explain your drawing.

Mathematics
2 answers:
lutik1710 [3]3 years ago
8 0
It's a square, parallelogram, or rhombus, because all three have 4 sides and corners and none of them are rectangles.
bija089 [108]3 years ago
4 0
Its a square that's the only other thing it could be ur welcome fammm
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Veronica’s velocity was measured as 4.3 m/s. She displaced 20 meters in 4.7 seconds. Which piece of information is missing for t
telo118 [61]

We are given displacement = 20 meters.

Time given = 4.7 seconds.

Velocity = 4.3 m/s.

Please note : Velocity is a vector quantity. A vector quantity depends on magnitude and direction as well.

We are given magnitude of displacement = 20 meter and time 4.7 seconds but the direction is missing there.

If it would be in forward direction it would be positive and if it would be in backward direction, it would be of negative.

Therefore, <u>"B)the direction"</u> is correct option.

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The answer is just 1/16
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Add using a number line. -3/4 + 1 1/4
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Find 1 1/4 on the number line and count back 3/4 (why it is negative) answer is 1/2.
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Hey im having trouble understanding 40 divided by 8? could you help me out haha
meriva

Answer:

40/8

Step-by-step explanation:

40 divided by 8 is 5.

The way you can check this is by multiplying 8*5

Basically what number times 8 will equal 40

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What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
3 years ago
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