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a_sh-v [17]
3 years ago
6

Write y =3/5 x + 3 in standard form using integers

Mathematics
1 answer:
liraira [26]3 years ago
5 0
It is already simplified

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Abby can buy an 8-pound bag of dog food for $7.40 or a 4- pound of the same dog food for 5.38. Which is better buy?
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The answer to your Q the better one it is the 4- pound bag because it coast less money
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What is the equation of this line in standard form?
Fiesta28 [93]

keeping in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{\frac{1}{2}}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{\frac{1}{2}}-\underset{x_1}{(-3)}}}\implies \cfrac{2+1}{\frac{1}{2}+3}\implies \cfrac{3}{\frac{1+6}{2}}\implies \cfrac{\frac{3}{1}}{~~\frac{7}{2}~~}\implies \cfrac{3}{1}\cdot \cfrac{2}{7}\implies \cfrac{6}{7}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{6}{7}}[x-\stackrel{x_1}{(-3)}]\implies y+1=\cfrac{6}{7}(x+3) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{7}}{7(y+1)=7\left( \cfrac{6}{7}(x+3) \right)}\implies 7y+7=6(x+3)\implies 7y+7=6x+18 \\\\\\ 7y=6x+11\implies -6x+7y=11\implies 6x-7y=-11

3 0
3 years ago
Which statement describes a process to solve sqrt b+20 - sqrt b=5?
babunello [35]

Answer:

Step-by-step explanation:

The equation is:

√b+20 - √b = 5

The first step is we will add √b to both sides:

√b+20 -√b +√b = 5 +√b

√b+20 = 5+√b

Now take square at both sides:

(√b+20)^2 = (5+√b)^2

b+20 = 25+10√b+b

Now combine the like terms:

b+20-25-b=10√b

-5 = 10√b

Divide both the terms by 10

-5/10 = 10√b/10

-1/2=√b

Take square at both sides:

(-1/2)^2 = (√b)^2

1/4 = b

So in this type of question we add radical terms to both sides and square both sides twice....

7 0
3 years ago
How do I simplify this (sinx)/(secx+1)
pochemuha

Answer:

Step-by-step explanation:

\frac{sin~x}{sec~x+1} =\frac{sin~x}{\frac{1}{cos~x}+1  } =\frac{sin~x~cos~x}{1+cos~x} \\=\frac{sin~x~cos~x}{1+cos~x} \times \frac{1-cos~x}{1-cos~x} \\=\frac{sin~x~cos~x(1-cos~x)}{1-cos^2 x} \\=\frac{sin~x~cos~x(1-cos~x)}{sin^2x} \\=\frac{cos~x(1-cos~x)}{sin~x} \\=cos ~x(csc~x-cot~x)\\or\\=cot~x(1-cos~x)

6 0
3 years ago
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