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2 years ago

Please help SUMMER SCHOOL

Mathematics

1 answer:

mote1985 [20]2 years ago

7 0

I am not sure what you mean by slope but the y intercept is (0,2) and the gradient is 3 making the equation y=3x+2
Explanation:
The y-intercept is where the line crosses the y axis and I have put the co-ordinate where it does that.
The gradient is 3 as to find the gradient you do rise/run so we do 3/1 which equals 3
To form an equation we do y=Mx+c. The c is the y-intercept and the gradient is m. Therefore the equation of this line is y= 3x+2.

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Yellow cab taxi charges $3.00 flat rate and $0.65 per mile. If Katie has no more than $10 to spend. How many miles can she trave

Alekssandra [29.7K]

Answer:

0.65+0.65=1.30+0.65=1.95+0.65=2.60

Step-by-step explanation:

well katie if katie wants to spend 3.00 she would have to do add it 4 times and it will be 2.60 and she would save 0.40 cents but if she adds it again it would be 3.25.

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3 years ago

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3 years ago

What is the measure of arc bc? A. 76 B. 40 C. 96 D. 58

Veseljchak [2.6K]

Answer:

A. 76

Step-by-step explanation:

Angle A is half the difference between the measures of arc DE and BC.

m∠A = (1/2)(DE -BC)

20 = (1/2)(116 -BC) . . . . substitute the given values

40 = 116 -BC . . . . . . . . multiply by 2

BC = 116 -40 . . . . . . . . add BC -40

BC = 76

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Comment on intersecting secants

When the secants intersect inside the circle, the angle where they cross is half the sum of the intercepted arcs. When they intersect outside the circle (as here), the angle where they meet is half the difference of the intercepted acs.

Sometimes it is easier to remember two related relationships than it is to remember just one of them.

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3 years ago

What is the mean of the data set in the box below? -8, -6, - 2, 0, -6, - 8. 4, 7, 8, 1

Andreas93 [3]

The mean of the data set in the box below is -1

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3 years ago

CALC- limits please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

2 years ago