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fenix001 [56]
3 years ago
12

What is the answer NEED HELP ASAP b-14=9 WHO ever answers gets 10 points

Mathematics
2 answers:
SpyIntel [72]3 years ago
5 0

Answer:

you only have 5 points or 10 points so whoever answers gets 5 points

also, b=23

Step-by-step explanation:

Naddik [55]3 years ago
4 0

Answer:

b-14 = 9

b = 23

Step-by-step explanation:

Add 14 to 9 to get b alone.

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What is the length of ef in the right triangle below?
katrin2010 [14]

Answer: A

Step-by-step explanation:

Using pythagoras;

25 = \sqrt{7^{2}+ x^{2} }\\25^{2}=7^{2} +x^{2}   \\625 = 49 +x^{2} \\576 = x^{2} \\x = 24

Please give me a brainliest answer

3 0
3 years ago
Read 2 more answers
What is the value of sin-1(1)? negative startfraction pi over 2 endfraction 0 startfraction pi over 2 endfraction π
mariarad [96]

The value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.

<h3>What is the value of sin (π/2)?</h3>

The value of sin (π/2) is equal to the number 1. The value of the sin-1(1) has to be find out.

Suppose the value of this function is <em>x</em>. Thus,

x=\sin^{-1}(1)

Solve it further,

\sin(x)=(1)              ......1

The value of sin (π/2) and -sin (-π/2) is equal to 1 such that,

\sin\dfrac{\pi}{2}=-\sin\left(-\dfrac{\pi}{2}\right)=1

Put this value in the equation 1,

\sin(x)=\sin\left(\dfrac{\pi}{2}\right)=-\sin\left(-\dfrac{\pi}{2}\right)

Thus, the range will be,

\left(\dfrac{\pi}{2},-\dfrac{\pi}{2}\right)

Thus, the value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.

Learn more about the sine values here;

brainly.com/question/10711389

3 0
2 years ago
F(x) = 2/3 x -5 what is the value f(6)?
katovenus [111]
F(6)=2/3(6)-5
f(6)=4-5
f(6)=-1

Final answer: f(6)= -1
5 0
3 years ago
If 6 is a factor of a number, what other numbers must be factors of<br> the number?
zysi [14]

Answer:

2 and 3

Step-by-step explanation:

6 = 2 × 3, so if a number is divisible by 6, it is also divisible by 2 and 3.

8 0
3 years ago
About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive
Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that -r is not irrational, then -r is rational. Since -1 is rational, then (-1)(-r)=r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

3 0
2 years ago
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