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3 years ago

Which figure has the same area as the parallelogram? Will give brainliest if correct

Mathematics

1 answer:

Ber [7]3 years ago

6 0

It should be D since the area of parallelogram is baseXheight, which is 40.2X21.8.
the area of the rectangle in option D is also 40.2X21.8

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Jackie surveyed the first 20 students in the lunch line.

Yanka [14]

Answer:

Yes

Step-by-step explanation:

It is a random sample of students because they are just random people that Jackie is asking, its not only people in on grade, or only people on the basketball team.

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3 years ago

Read 2 more answers

Identify the type of polar graph for the equation: r = 3-5cos θ aLimacon with inner loop bCardioid cDimpled limacon dConvex lima

svp [43]

Given the equation:

$r=3-5\cos \theta$

Let's identify the type of polar graph for the equation.

To identify the type of polar graph, use the formula below to get the Cartesian form:

$(x^2_{}+y^2)=r(\cos \theta,\sin \theta)$

Thus, we have:

$(x^2+y^2)=3\sqrt[]{x^2+y^2}-5x$

We have the graph of the equation below:

We can see the graph forms a Limacon with an inner loop.

Therefore, the type of polar graph for the given equation is a limacon with inner loop.

ANSWER:

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1 year ago

Describe diagram....

sweet [91]

Answer:

(B ^ C) U (A U B UC)'

4 0

4 years ago

IF THE PICTURES ARE NOT THERE RELOAD THE PAGE BEFORE SAYING SOMETHING

Lelechka [254]

A is correct due to the smaller interquartile range

4 0

3 years ago

Write the equation for a 3rd degree polynomial function that has roots at x = 1–3i and x = 2. The y‑intercept is (0,10). Write a

HACTEHA [7]

Answer:

$\displaystyle P(x)=-\frac{1}{2}(x^2+2x+10)(x-2)$

Step-by-step explanation:

<u>Factored Form Of Polynomials </u>

If we know the roots of a polynomial as

$\alpha _1,\alpha _2,\alpha _3$

the polynomial can be expressed in factored form as

$a(x-\alpha _1)(x-\alpha _2)(x-\alpha _3)$

We are given two of the three roots of the polynomial:

$\alpha _1=1-3i$

$\alpha _2=2$

The other root must be the conjugate of the complex root:

$\alpha _3=1+3i$

Recall the product of two complex conjugate numbers is

$(1-3i)(1+3i)=1^2-(3i)^2=1+9=10$

The required polynomial is

$P(x)=a\left[ x-(1-3i)\right ]\left[ x-(1+3i)\right ](x-2)$

$P(x)=a(x^2+2x+10)(x-2)$

This is the factored form of the polynomial where only real numbers appear

We need to find the value of a, such as

$P(0)=10$

$P(0)=a(0^2+2(0)+10)(0-2)=10$

$-20a=10$

Thus the value of a is

$\displaystyle a=-\frac{1}{2}$

The expression of the required polynomial is

$\boxed{P(x)=-\frac{1}{2}(x^2+2x+10)(x-2)}$

5 0

3 years ago