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3 years ago

NEED ANSWER QUICK!!!!!!!

Mathematics

2 answers:

bija089 [108]3 years ago

6 0

-6(2) = -12

-5 7/8 is close to -6

2 4 /47 is close to 2

-12 1/4 is close to -12

-6 * 2 = -12

grigory [225]3 years ago

3 0

Answer:

bro idek but thx for points

Step-by-step explanation:

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1. Simplify 2/radical 5

cupoosta [38]

1. (2√5)/5
2. -44√7
3. 8√17
4. -6√3+12

For 1: Multiply the numerator and denominator by √5:
$\frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

For 2: Simplify √112:
√112 = √(2*56) = √(2*2*28) = √(4*4*7) = 2*2√7 = 4√7
Now multiply by the -11 coefficient:
-11(4√7) = -44√7

For 3: Add the radicals as you would variables that are like terms.

For 4: Multiply by the conjugate. The conjugate of the denominator has the same values with the sign of the radical changed:
$\frac{6}{\sqrt{3}+2}\times \frac{\sqrt{3}-2}{\sqrt{3}-2}=\frac{6(\sqrt{3}-2)}{(\sqrt{3}+2)(\sqrt{3}-2)}
\\
\\=\frac{6\sqrt{3}-6*2}{\sqrt{3}*\sqrt{3}-2\sqrt{3}+2\sqrt{3}-2*2}
\\
\\=\frac{6\sqrt{3}-12}{3-4}=\frac{6\sqrt{3}-12}{-1}=-6\sqrt{3}+12$

7 0

3 years ago

Can someone help me with this question?

solniwko [45]

The function is open up for both since the stretch value is positive. When opening you we don’t have a maximum value only a minimum value. So already we know B and d are wrong. When looking at minimums and maximums we look at the y or q value. In this case for f(x) it is 4 and for g(x) it is 5. So we know g(x) have a greater minimum value, so the answer is c

5 0

3 years ago

For the first one the answer are

BigorU [14]

Answer:

1. add 1/2x to both sides

a. you want to combine the like terms. in this case, it is the x variable.

you are left with 7/6x = 5

2. multiply by 6/7

a. the reciprocal of 7/6 will cancel out the values

4 0

3 years ago

Determine whether each of these sets is nite, countably in nite, or uncount- able. For those that are countably in nite, exhibit

ira [324]

Answer:

Integers that are divisible by 7 are countably infinite

The set are {7, 14, 21, 28, 35, 42, 49,......

Integers that are divisible by 10 are countably infinite

The set are {10, 20, 30, 40, 50, 60, 70,........

Also for the negative integers divisible by 7 or 10{.......,-49, -42, -35, -28, -21, -14, -7} or {........,-70, -60, -50, -40, -30, -20, -10} are countably infinite

Step-by-step explanation:

See the set of integers divisible by 7 or 10,

{7, 14, 21, 28, 35, 42, 49,........} Or

{10, 20, 30, 40, 50, 60, 70,.......}

Can be map one-to-one to {1,2,3,4,5,6,7,...... The set of natural numbers or positive Integers.

So also, it is applicable in negative integers divisible by 7 or 10.

Therefore, they are countably infinite

A set is uncountable if it contains so many elements that they cannot be put in one-to-one correspondence with the set of natural numbers(Positive Integers). In other words, there is no way that one can count off all elements in the set in such a way that, even though the counting will take forever, you will get to any particular element in a finite amount of time.

WHILE

A set is countable or countably finite if it contains so many elements that they CAN be put in one-to-one correspondence with the set of natural numbers(Positive Integers)

3 0

3 years ago

A line parallel to y=1/3 x+2

Kamila [148]

Answer:

y = 1/3x

Step-by-step explanation:

If you use the form y = mx + b, all you need to create a parallel line is to have the same slope. So, we can just disregard b and write an equation with a slope of 1/3. You can add or subtract any number to that if you want.

3 0

3 years ago