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3 years ago

What is the value of the expression 3y−zy+5z3y−zy+5z when y = 6 and z = 2?

Mathematics

2 answers:

WARRIOR [948]3 years ago

6 0

The valve of the expression would be 184

Vinvika [58]3 years ago

5 0

Plug in the numbers for y and z. 3(6)-6(2)+5(2). Solve. 18-12+(10). The answer is 16.

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What is the product of 3/4 and -6/7 ?

Allisa [31]

Answer:

-9 / 14

Step-by-step explanation:

Product means multiply
3/4 x -6/7 = -9 / 14

7 0

3 years ago

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HELP PLEASE WILL LEAVE GOOD RATING

Arisa [49]

Answer:

that would be 200 to 2,000 feet in the air

3 0

3 years ago

The points -4. -4 -4, 4

Ilia_Sergeevich [38]

Answer:

whats the question?

Step-by-step explanation:

I don't get the question so what is the question?

3 0

3 years ago

PLEASE HELP ME IM STRUGGLING!!!

Kitty [74]

Answer:

The required answer is $c=7\sqrt{3}$

Therefore the number in green box should be 7.

Step-by-step explanation:

Given:

AB = 7√2

AD = a , BD = b , DC = c , AC = d

∠B = 45°, ∠C = 30°

To Find:

c = ?

Solution:

In Right Angle Triangle ABD Sine identity we have

$\sin B = \dfrac{\textrm{side opposite to angle B}}{Hypotenuse}\\$

Substituting the values we get

$\sin 45 = \dfrac{AD}{AB}= \dfrac{a}{7\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}= \dfrac{a}{7\sqrt{2}}\\\\\therefore a=7$

Now in Triangle ADC Tangent identity we have

$\tan C = \dfrac{\textrm{side opposite to angle C}}{\textrm{side adjacent to angle C}}$

Substituting the values we get

$\tan 30 = \dfrac{AD}{DC}= \dfrac{a}{c}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{7}{c}\\\\\therefore c=7\sqrt{3}$

The required answer is $c=7\sqrt{3}$

8 0

3 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

3 years ago