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iVinArrow [24]
3 years ago
11

What is the area of the figure in square millimeters and in square centimeters.

Mathematics
1 answer:
Westkost [7]3 years ago
5 0
Area 15x15= 225 mm^2
Area 10X10=100mm^2
Total 325 mm^2 or 3.25 cm^2
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1. what is the equation of a line that has an x-intercept of 2 and a y-intercept of 8
Phoenix [80]

Answer:

B

Step-by-step explanation:

x- intercept = 2  .So, Coordinate(2 , 0)

y-intercept = 8 . So, Coordinate (0, 8)

Slope = \frac{8-0}{0-2}\\\\=\frac{8}{-2}\\\\=-4

Slope y-intercept form: y = mx + b  {Here, b is y-intercept}

y = -4x + 8

8 0
2 years ago
Pls :,) i suck at math
vesna_86 [32]

Answer:

60%

Step-by-step explanation:

The original was $20 to $12, so if you divide both prices by 2 (to make one side 10 and the other the simplified percentage) you get:

10 to 6.

Percentages are typically x% out of 100, so if you multiply both by 10 you get:

100 to 60. All you have to do now is change the format to percentage instead of a fraction (60/100) and the answer is 60%.

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6 0
2 years ago
Read 2 more answers
There are 20 dowels. All bevels are dowels. All dowels are swivels. There are 12 bevels. There are 30 swivels. How many swivels
Zina [86]

Answer:

10 swivels are neither bevels or dowels

Step-by-step explanation:

See the attached for explanation

7 0
3 years ago
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Dont answer if you dont have the answer I will report you The graph of a function contains the points (–1, –4), (0, 1), (1, 4).
Fantom [35]

Answer:

  • The function is linear because the points are not in a straight line.

Points that lie along the same line are said to be "collinear" and collinear points must have the same slope between any pair of points.

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4 0
3 years ago
Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\
motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in \sqrt{7}, 1 - \sqrt{6} and -3. The last root gives us the factor (x+3). Hence, our polynomial is

P(x) =(x+3)q(x)

where q is a polynomial with rational coefficients and roots \sqrt{7} and 1 - \sqrt{6}. The root \sqrt{7} gives us a factor x-\sqrt{7}, but in order to obtain rational coefficients we must consider the factor x^2-7.

An analogue idea works with 1 - \sqrt{6}. For convenience write  x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}. This gives the factor (x-1)^2-6. Hence,

P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105

Notice that P(-1)=24. So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type x-\sqrt7 will introduce in the expression, we need to multiply by its conjugate x+\sqrt7. Hence, we will obtain x^2-7 that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0
2 years ago
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