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3 years ago

What is the best first step to find the slope and y-intercept when given an equation of a line?

Mathematics

2 answers:

Llana [10]3 years ago

5 0

In y=mx+c, c is the y-intercept or make x=0.

stiv31 [10]3 years ago

5 0

When given a slope intercept form type of equation (y= mx + b), automatically, the value of m, which is the coefficient of x is the slope and the value of b, which is also known as the constant is the y intercept

But when given another type of form in an equation, always isolate y first for you to get a value of m and b

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Help me, please ...i dont understand

BARSIC [14]

C. Use the formula for area of a triangle 1/2(Base * Height). Using that we see that we have 1/2(a*b) + 1/2(c*c) + 1/2(a*b). Cleaning this up we get 1/2(2ab+c^2)

3 0

3 years ago

Austin takes 4 hours to make 2 backpacks and a handbag. The time he takes to

lys-0071 [83]

Step-by-step explanation

b = 2h is given

2 b + h = 4 hours is given sub in the top equation

2 (2h) + h = 4

5h = 4

h = 4/5 hr or 48 minutes

7 0

3 years ago

the first term of an arithmetic sequence is 10 and the sum of the first five terms is 250,find the sequence

Crazy boy [7]

Answer:

The sequence is:

10, 30, 50, 70, 90.....................

Step-by-step explanation:

We have,

First term (a) = 10

Common difference (d) = ?

Sum of first 5 terms ( $S_{5}$ ) = 250

or, $\frac{n}{2} [{2a+(n-1)d}] = 250$

or, $\frac{5}{2} [2*10 + 4d]=250$

or, $\frac{5}{2} * 4[5+d]=250$

or, 10(5 + d) =250

or, 5 + d = 25

∴ d = 20

Now,

2nd term = a + d = 10 + 20 = 30

3rd term = a + 2d = 10 + 2*20 = 10 + 40 = 50

4th term = a + 3d = 10 + 3*20 = 10 + 60 = 70

5th term = a + 4d = 10 + 4*20 = 10 + 80 = 90

8 0

3 years ago

Gerald has a punch bowl in the shape of a hemisphere as shown below.

zmey [24]

The closest to the maximum number of cups the punch bowl can hold is 30

<h3>How to determine the number of cups?</h3>

The given parameters are:

1 cup = 15 cubic inches

Diameter of bowl, d = 12 inches

The radius is the half of the diameter.

So, we have:

r = 6

The volume of the bowl is then calculated using:

$V = \frac{2}{3}\pi r^3$

This gives

$V = \frac{2}{3}\pi * 6^3$

Evaluate

V = 452.16

The maximum number of cups is then calculated using:

Cups = 452.16/15

Evaluate

Cups = 30.1444

Approximate

Cups = 30

Hence, the closest to the maximum number of cups the punch bowl can hold is 30

Read more about volumes at:

brainly.com/question/1972490

#SPJ1

6 0

2 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago