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2 years ago

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Your current test average in Science is an 81. You’ve taken 5 tests. You only have one test remaining (there is not extra credit

on your test). You can raise your average to an 85.

2 answers:

stepladder [879]2 years ago

8 0

If you want to know what grade is needed to raise your average to an 85, add all tests, then divide by total number of tests to equal average of 85. The x value is the grade needed on the final to raise the overall class grade to an 85.

x= final test

(81 + 81 + 81 + 81 + 81 + x)/6= 85
add in parentheses

(405 + x)/6= 85
multiply both sides by 6

405 + x= 510
subtract 405 from both sides

x= 105 grade needed

ANSWER: Since there is no extra credit, it is impossible to raise the overall average to an 85 because they would need a final test grade of 105.

Hope this helps! :)

Leto [7]2 years ago

5 0

What exactly are you asking?

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A book has 89 pages, but the page numbers are printed incorrectly. Every third page number has been omitted, so that the pages a

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Answer:

89

Step-by-step explanation:

There are only 30 pages but last page number is 89.

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1 year ago

Find all of the polar coordinates of point P if p=(1,-pi/6)

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Answer:

The all polar coordinates of point P are (1 , -pi/6 + 2nn) and (-1, -pi/6+(2n+1)n).

Step-by-step explanation:

The polar coordinates can be written as (r,Ф)= (r,Ф+2nn) or (r,Ф) =(-r,Ф+(2n+1)n) and n is any integer value.

We are given p=(1,-pi/6)

P(r,Ф), then r = 1 and Ф = -pi/6

The polar coordinates will be

P(r,Ф)= (r,Ф+2nn) = (1 , -pi/6 + 2nn) and n is any positive integer and the value of r is positive.

P(r,Ф) =(-r,Ф+(2n+1)n) = (-1, -pi/6+(2n+1)n) and n is any positive integer and the value of r is negative.

The all polar coordinates of point P are (1 , -pi/6 + 2nn) and (-1, -pi/6+(2n+1)n)

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1 year ago

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Define a function f on a set of real numbers as follows: f(x) = 3x − 1 x , for each real number x ≠ 0 Prove that f is one-to-one

dalvyx [7]

Answer with Step-by-step explanation:

We are given that

$f(x)=\frac{3x-1}{x}$

For each real number $x\neq 0$

To prove that f is one -to-one.

Proof:Let $x_1$ and $x_2$ be any nonzero real numbers such that

$f(x_1)=f(x_2)$

By using the definition of f to rewrite the left hand side of this equation

$f(x_1)=\frac{3x_1-1}{x_1}$

Then, by using the definition of f to rewrite the right hand side of this equation of $f(x_1)=f(x_2)$

$f(x_2)=\frac{3x_2-1}{x_2}$

Equating the expression then we get

$\frac{3x_1-1}{x_1}=\frac{3x_2-1}{x_2}$

$3x_1x_2-x_2=3x_2x_1-x_1$

$3x_1x_2-3x_1x_2+x_1=x_2$

$x_1=x_2$

Therefore, f is one-to-one.

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1 year ago

Which statements about the figure are true? Select two options.

harina [27]

<u>Complete Question</u>

The circle is inscribed in triangle PRT. A circle is inscribed in triangle P R T. Points Q, S, and U of the circle are on the sides of the triangle. Point Q is on side P R, point S is on side R T, and point U is on side P T. The length of R S is 5, the length of P U is 8, and the length of U T is 6. Which statements about the figure are true?

Answer:

(B)TU ≅ TS

(D)The length of line segment PR is 13 units.

Step-by-step explanation:

The diagram of the question is drawn for more understanding,

The theorem applied to this problem is that of tangents. All tangents drawn to a circle from the same point are equal.

Therefore:

|PQ|=|PU|=8 Units

|ST|=|UT| =6 Units

|RS|=|RQ|=5 Units

(b)From the above, TU ≅ TS

(d)Line Segment |PR|=|PQ|+|QR|=8+5=`13 Units

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2 years ago

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Answer:

The distance around the lot is 4.05 miles .

Step-by-step explanation:

Formula

$Perimeter\ of\ rectangle = 2 (Length + Breadth)$

As given

$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ 1 \frac{1}{8}\ miles\ by\ \frac{9}{10}\ mile.$

i.e

$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ \frac{9}{8}\ miles\ by\ \frac{9}{10}\ mile.$

Here

$Length = \frac{9}{8}\ miles$

$Breadth = \frac{9}{10}\ miles$

Put in the formula

$Perimeter\ of\ rectangle = 2(\frac{9}{8}+\frac{9}{10})$

L.C.M of (8,10) = 40

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$Perimeter\ of\ rectangle = \frac{2\times (45+36)}{40}$

$Perimeter\ of\ rectangle = \frac{2\times 81}{40}$

$Perimeter\ of\ rectangle = \frac{81}{20}$

Perimeter of rectangle = 4.05 miles

Therefore the distance around the lot is 4.05 miles .

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2 years ago

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