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NeTakaya
3 years ago
8

jubal wrote the four equations below. He examined them, without solving them, to determine which equation had no solution

Mathematics
2 answers:
denis-greek [22]3 years ago
7 0
The answer to the question is b
Murljashka [212]3 years ago
3 0

Answer:

It's B I took the test


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Lisa needed 1 cup of white sugar to 2 cups of brown sugar to make one batch of cookies.What would be the ratio of white sugar to
raketka [301]
1 cup : 2 cups of brown sugar

1:2

4 batches of cookies, so multiply both numbers by 4. 

4:8

Answer: 4 to 8
4 0
3 years ago
In this diagram, BAC – EDF. If the
puteri [66]
The triangles are congruent. So.

Triangle ABC is 24 inch^2 and BC is 4 inches.
The area of a triangle is 0.5 x base x height. The will be 12 because 12 x 4 x 0.5 which equals 24.

Triangle DEF is half the size of ABC because bc is 4 and EF is 2 showing it’s half. This then means the height will be half which then mean it will be 6.

This then means we can use the equation to work it out. So. 6 x 0.5 x 2 = 6in^2

The answer is 6inches^2
6 0
3 years ago
What is the measure of Angle y in this figure?
uranmaximum [27]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here, in the given figure. Angle y° forms vertical opposite angle pair with Angle 163°, therefore they have equal values.

that means ~

\qquad \sf  \dashrightarrow \: y \degree = 163 \degree

3 0
2 years ago
Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28. refer to exhibit 6-
tatyana61 [14]
(25 -21)/(28 -20) = 4/8 = 1/2

Your probability is 1/2.
7 0
3 years ago
Find a polynomial function of degree 3 with 2, i, -i as zeros.
sergiy2304 [10]

Answer:

p(x)= x^3-2x^2+x-2

Step-by-step explanation:

Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if \alpha , \ \beta \ \& \ \gamma are the zeros of the cubic polynomial , then ,

\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma)

Here in place of the Greek letters , substitute 2,i and -i , we get ,

\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i)

Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,

\sf  \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\}

Simplify using i = √-1 ,

\sf \longrightarrow p(x)= (x-2)( x^2 + 1 )

Multiply by distribution ,

\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1)

Simplify by opening the brackets ,

\sf\longrightarrow p(x)= x^3+x-2x^2-2

Rearrange ,

\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}

4 0
2 years ago
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