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Elenna [48]
2 years ago
13

Solve this:

Mathematics
1 answer:
Alex73 [517]2 years ago
7 0

Answer:

i = 17 Years old
ii = 25 years

Step-by-step explanation:

i = 12 + 5 =17

ii = his son was born when he was 9, hes 12 now, so his sons 3. In 5 years time his sons 8 and hes 17, 8+17= 25

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Unit 4 Test, Part 2: Congruence and Constructions<br> Help with my math please
lakkis [162]

Answer/Step-by-sep explanation:

1. Given:

∆NMK ≅ ∆TRP

\overline{NM} = 20

\overline{MK} = 15

\overline{KN} = 25

\overline{TR} = 3x - 1

a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP

b. The side that is congruent to \overline{TR} is \overline{NM}. Thus:

\overline{TR} ≅ \overline{NM}

c. Since \overline{TR} ≅ \overline{NM}, therefore:

\overline{TR} = \overline{NM}

3x - 1 = 20 (substitution)

Add 1 to both sides

3x = 20 + 1

3x = 21

Divide both sides by 3

x = \frac{21}{3}

x = 7

2. a. Slope of LK = \frac{rise}{run} = \frac{4}{3}

Slope of LM = \frac{rise}{run} = -\frac{3}{5}

b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):

Lk = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2}

Lk = \sqrt{(3)^2 + (4)^2}

Lk = \sqrt{9 + 16}

Lk = \sqrt{25}

Lk = 5

✍️Length of LM is the distance between L(-7, 4) and (-2, 1):

LM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2}

LM = \sqrt{(5)^2 + (-3)^2}

LM = \sqrt{25 + 9}

LM = \sqrt{34}

LM = 5.8 (nearest tenth)

∆KLM is not an isosceles ∆ because it does not has two equal side lengths. This we can see because LK and LM are not equal.

Therefore, Anthony is incorrect. Am isosceles ∆ has two equal sides.

8 0
3 years ago
What are the eletric charges of three subatomic particles​
Nata [24]

Answer:

3

Step-by-step explanation:

as a person who answers in the mathematics session

i will tell you that the answer is 3

7 0
3 years ago
Translate and solve: 639 is what percent of 142?
Andre45 [30]

Answer:

450

Step-by-step explanation:

639 is what percent of 142? = 450.

4 0
3 years ago
Read 2 more answers
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
juin [17]

Answer:

The Taylor series of f(x) around the point a, can be written as:

f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....

Here we have:

f(x) = 4*cos(x)

a = 7*pi

then, let's calculate each part:

f(a) = 4*cos(7*pi) = -4

df/dx = -4*sin(x)

(df/dx)(a) = -4*sin(7*pi) = 0

(d^2f)/(dx^2) = -4*cos(x)

(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4

Here we already can see two things:

the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.

so we only will work with the even powers of the series:

f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....

So we can write it as:

f(x) = ∑fₙ

Such that the n-th term can written as:

fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}

6 0
3 years ago
Stephanie and victor each built a rectangular garden .Stephanies garden is three times as long and tree times as wide as victori
MrRa [10]

Answer:multiply by three

Step-by-step explanation:

8 0
3 years ago
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