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3 years ago

Can Someone Help me with my Algabra 2 Homework 4a-c

Mathematics

1 answer:

kicyunya [14]3 years ago

6 0

Answer:

Step-by-step explanation:

Sigma notation is:

∑ ak

k=1

The n on top is the number of terms. ak is the expression for the kth term.

Let's look at the first one. The series is:

-2+4+-8+...+64+-128

There's two things to notice. One, the sign changes back and forth between + and -. Two, the magnitude doubles with each next term. Therefore:

ak = (-2)ᵏ

Next we need to find the number of terms. -128 is the last term, so:

128 = 2ⁿ

n = 7

So the answer is:

∑ (-2)^k

k=1

Now the second one. Notice the numerators are all 1 and the denominators are all perfect squares. Therefore:

ak = (1/k)²

The last term is 1/100, so n = 10.

So the answer is:

∑ (1/k)²

k=1

Now the last one. Notice that each term is 5 plus the previous term. This is an arithmetic series. So we can say:

ak = 4 + 5(k-1)

ak = 4 + 5k - 5

ak = 5k - 1

The last term is 49, so:

49 = 5n - 1

n = 10

So the answer is:

∑ (5n-1)

k=1

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Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.

dem82 [27]

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)

Formula used :

→Mid point of two points- $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

→Slope of two points - $\frac{y_2-y_1}{x_2-x_1})$

→Perpendicular of a line = $\frac{-1}{slope of line})$

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = -1

Equation of AB with slope -1 and the coordinates (-1,0) is,

(y – 0) = -1(x – (-1))

y+x=-1………………(1)

Similarly, for AC

Mid point of AC = $(\frac{2+2}{2},\frac{3-3}{2})=(2,0)$

Slope of AC = $\frac{-3-3}{2-2}=\frac{-6}{0}$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 0

Equation of AC with slope 0 and the coordinates (2,0) is,

(y – 0) = 0(x – 2)

y=0 ………………(2)

By solving equation (1) and (2),

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of CF = -1

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF

(y – 0) = -1(x – (-1))

y+x=-1………………(3)

Similarly, mid point of BC = $(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)$

Slope of AB = $\frac{-3+3}{-4-2}=0$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))

y+3=0

y=-3………………(4)

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)

7 0

3 years ago

Dina wants to make 15 3/4 cups of strawberry drink by mixing water and strawberry syrup with a ratio of 2 1/4 cup of water for e

bekas [8.4K]

Answer:

The water she need is $7\frac{7}{8} \ cups.$ and strawberry syrup $11\frac{13}{16}\ cups$ .

Step-by-step explanation:

Given:

Dina wants to make 15 3/4 cups of strawberry drink by mixing water and strawberry syrup with a ratio of 2 1/4 cup of water for every 3/4 cup of syrup.

Now, to find the quantity of water and syrup she need to use.

As given in question ratio so:

Strawberry syrup = 2 1/4 = 9/4.

Water = 3/4.

Total cups of strawberry drink = 15 3/4 = 63/4.

Let the strawberry syrup be $\frac{9}{4} x$ .

And let the water be $\frac{3}{4} x$ .

According to question:

$\frac{9x}{4} + \frac{3x}{4}=\frac{63}{4}$ .

On adding the fractions:

⇒ $\frac{9x+3x}{4} =\frac{63}{4}$

⇒ $\frac{12x}{4} =\frac{63}{4}$

Multiplying both sides by 4 we get:

⇒ $12x=63$

Dividing both sides by 12 we get:

⇒ $x=\frac{63}{12}$

Dividing numerator and denominator by 3 on R.H.S we get:

⇒ $x=\frac{21}{4}$

Now, putting the value of $x$ on ratios:

Strawberry syrup = $\frac{9}{4}\times x=\frac{9}{4}\times\frac{21}{4}$

= $\frac{189}{16}$

= $11\frac{13}{16}\ cups$

Water = $\frac{3}{4}\times x =\frac{3}{4} \times\frac{21}{4}$

= $\frac{63}{8}$

= $7\frac{7}{8} \ cups.$

Therefore, the water she need is $7\frac{7}{8} \ cups.$ and strawberry syrup $11\frac{13}{16}\ cups$ .

4 0

3 years ago

your phone says that 14.5% is equivalent to the decimal 14.5 explain why your friend is incorrect by comparing the fraction -14.

Lelechka [254]

A percent is a proportion of something that can come from a variety of different numbers depending on what you are measuring. 14.5% is not the same as 14.5 because the 14.5% isn’t necessarily 14.5 of a certain amount. For example, $14.50 of the $100 you have could be from your grandparents; making 14.5% of your money from your grandparents but a deposit of $145 into a bank account that already has $855 would mean that 14.5% of the money in that bank account was from that deposit but $145 does not equal $14.50. They just have the same proportion just different on different scales.

Hope this is what you were looking for.

4 0

2 years ago

A number divided by 6 minus -3 equals 12

liberstina [14]

Let the number be x.

Therefore, by the problem

x/6-(-3) = 12

or, x/6+3 = 12

or, x/6 = 12-3

or, x/6 = 9

or, x = 9 × 6

or, x = 54

The number is 54.

4 0

2 years ago

Read 2 more answers

The equation of the parabola whose focus is at (7 0) and directrix at y=-7 is

bija089 [108]

The equation of the parabola whose focus is at (7, 0) and directrix at y = -7 is
(x-7)^2 = 14(y + 7/2)

Hope this helps

5 0

3 years ago

Can Someone Help me with my Algabra 2 Homework 4a-c​

Can Someone Help me with my Algabra 2 Homework 4a-c