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umka21 [38]
2 years ago
13

Which is the length of the third slide of the right triangle

Mathematics
1 answer:
irina [24]2 years ago
8 0

Answer:

27

Step-by-step explanation:

if you do pathagorean therom

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Emmy used one-centimeter cubes to build a larger cube. The edges of Emmy's cube are 3 centimeters long. Tim has a cube-shaped bo
butalik [34]

Answer:

oof

Step-by-step explanation:

8 0
2 years ago
Sam bought 3 pens for $0.49
Zigmanuir [339]

Answer:

The answer should be 3.15

Step-by-step explanation:

Multiply $0.49 by 3 and you get $1.47.

Add $1.59 and $1.47 and you get $2.97

Add $2.97 plus 6% and you get 3.1482

$3.15.

3 0
3 years ago
jacks tent is shaped like an isosceles triangle. the height of the tent measures 7 feet and the diagonal side measures 9 feet. f
abruzzese [7]

Answer:

8\sqrt{2}\ ft

Step-by-step explanation:

see the attached figure to better understand the problem

Remember that

An isosceles triangle has two equal sides and two equal interior angles

In the isosceles triangle ABC

Applying the Pythagorean Theorem

Let

b ----> the length of the tents base

9^2=7^2+(b/2)^2

(b/2)^2=9^2-7^2

(b^2/4)=32

b^2=128\\\\b=\sqrt{128}\ ft

simplify

b=8\sqrt{2}\ ft

3 0
3 years ago
Can someone help me with this? <br> **SHOW WORK PLEASE**
Valentin [98]

FYI, that L in the denominator is factorial.

This is some pretty serious stuff for high school.

My inclination is that the answer is A, which is the definition of <em>e. </em> Let's see if we can show this.

Let's write

\displaystyle f(n) = \left(1 + \dfrac 1 n \right)^n

Let's expand this with the binomial expansion

\displaystyle f(n) =\sum_{k=0}^n {n \choose k} \dfrac{1}{n^k}

\displaystyle f(n) =\sum_{k=0}^n \dfrac{n!}{k!(n-k)!} \cdot \dfrac{1}{n^k}

\displaystyle f(n) =\sum_{k=0}^n \dfrac{n(n-1)\cdots(n-k+1)}{k! \, n^k}

Let's focus on when n is really big and on the ks that are relatively small, which make up the bulk of the sum as the terms get small rapidly.

Then that numerator n(n-1)···(n-k+1) ≈ n^k as all the factors are about n.  

\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{n^k}{k!\, n^k}

\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{1}{k!}

OK, we showed for large n this is approximately true, and it will be exactly true in the limit, so we choose

Answer: A

3 0
3 years ago
What system of linear inequalities is shown in the graph?
Oksanka [162]

Answer:  LINEAR PERSPECTIVE

5 0
3 years ago
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