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3 years ago

Solve the equation 3x+5y=16 for y

Mathematics

2 answers:

postnew [5]3 years ago

6 0

X= 2
Y=2
This hopefully is the answer

Nikitich [7]3 years ago

4 0

Y = 16/5 - 3/5x that's the answer

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Find the value of S for an infinite geometric series with r =1/6 and a1= 10.

fiasKO [112]

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3 years ago

Question 10

KonstantinChe [14]

Answer:

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3 years ago

Simplify (12a5−6a−10a3)−(10a−2a5−14a4) . Write the answer in standard form

almond37 [142]

Answer:

$=14a^5+14a^4-10a^3-16a$

Step-by-step explanation:

$\left(12a^5-6a-10a^3\right)-\left(10a-2a^5-14a^4\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=12a^5-6a-10a^3-\left(10a-2a^5-14a^4\right)\\-\left(10a-2a^5-14a^4\right):\quad -10a+2a^5+14a^4\\-\left(10a-2a^5-14a^4\right)\\\mathrm{Distribute\:parentheses}\\=-\left(10a\right)-\left(-2a^5\right)-\left(-14a^4\right)\\Apply\:minus-plus\:rules\\-\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\=-10a+2a^5+14a^4\\=12a^5-6a-10a^3-10a+2a^5+14a^4$

$\mathrm{Simplify}\:12a^5-6a-10a^3-10a+2a^5+14a^4:\quad 14a^5+14a^4-10a^3-16a12a^5-6a-10a^3-10a+2a^5+14a^4\\Group\:like\:terms\\=12a^5+2a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:12a^5+2a^5=14a^5\\=14a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:-6a-10a=-16a\\=14a^5+14a^4-10a^3-16a$

4 0

3 years ago

What is the value of x?

blsea [12.9K]

Since these are supplementary angles (two angles which sum to 180°) we can say:

4x+x=180 combine like terms on left side

5x=180 divide both sides by 5

x=36°

3 0

3 years ago

Can someone please do this for me?

san4es73 [151]

Answer:

64°

Step-by-step explanation:

So supplement angles are equal to 180, so first our two angles added together will be 180:

x + x = 180

Now one of these angles is 6 greater than half:

x + (x/2 + 6) = 180

Then we will solve for x:

x + (x/2 + 6) = 180

x + x/2 + 6 = 180

x + x/2 = 174

$\frac{3}{2}$ x = 174

x = 116

Plug it back in:

x + (x/2 + 6) = 180

x/2 + 6

(116)/2 + 6

To check:

116 + 64 = 180 ✓

6 0

3 years ago