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Reil [10]
3 years ago
11

What is the value of h?

Mathematics
1 answer:
elena-s [515]3 years ago
4 0

Answer:

B) h=8√2

Step-by-step explanation:

In a 45-45-90 triangle, h=s√2, where s is the side length of the triangle. Therefore, h=8√2

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What is the nth term of 4 1 -2 -5 -8
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The nth Term of this number is -3n+7

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THIS IS URGENT I NEED HELP ASAP.(9x 10^-3)^2
joja [24]

(9x 10^-3)^2=\frac{81x^{2}}{1000000}, so d 8.1x10^-6

Evaluate 1/2a^-4b^2 for a =-2 and b=4

\frac{1}{2} (-2)^{-4} *4^{2}\\\\\frac{1}{2} *\frac{1}{(-2)^{4}} *4^{2}\\\\\frac{1}{2} *\frac{1}{2^{4}}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*16=\frac{1}{2}

A number raised to a negative exponent is sometimes negative.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)  

- Cutiepatutie ☺❀❤

8 0
3 years ago
The product (x) of two numbers is 24 and their sum (+) is 10. What is the value of the largest of the two numbers?
RideAnS [48]

let the two numbers be x and y

From the first sentence,

xy=24

x+y=10

Then make y in equation 2 the subject of the formular and substitute in equation 1

x+y=10

y=10-x

substituting in equation 2

x(10-x)=24

open the bracket

10x-x^2=24

=-x^2+10x=24

Transfer the constant to the left hand side

=-x^2+10x-24=0

Then factorise completely

Look at the photo above

4 0
3 years ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that p = 0.95

10 targets

This means that n = 10

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

P(X = 10) + P(X < 10) = 1

We want P(X < 10). So

P(X < 10) = 1 - P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987

P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401

40.1% probability that he will miss at least one of them

7 0
3 years ago
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