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3 years ago

Solve the equation 4x^2 - 12x = 7 algebraically for x.

1 answer:

azamat3 years ago

3 0

$4x^2-12x=7\\4x^2-12x-7=0\\x=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)+-\sqrt{(-12)^2-4(4)(-7)}}{2(4)}\\x=\frac{12+-\sqrt{144+112}}{8}\\x=\frac{12+-\sqrt{256}}{8}\\x=\frac{12+-16}{8}\\x=\frac{-4,28}{8}\\x=\frac{-1}{2},\frac{7}{2}$

x = -0.5, 3.5

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

In-s [12.5K]

Answer:

593 1/3 OR 593.3 (the 3 is repeating)

Step-by-step explanation:

add all the numbers up then divide by the number of results (6)

3 0

4 years ago

Pls help I am stuck on this question

Elanso [62]

Answer:

153.94 cm

Step-by-step explanation:

a = r^pi squared

r = d/2

7^pi = 21.98

21.98 squared = 153.94 cm

5 0

3 years ago

Read 2 more answers

Haley has 3 3/8 cups of sugar to make pies. If each pie uses 2 1/4 cups of sugar, how many pies can she make?

tester [92]

Answer:

With 3 3/8 cups of sugar Haley can make one and a half pie.

Step-by-step explanation:

The amount of sugar required to make each pie = 2 1/4 cup

Now, $2 \frac{1}{4} = 2 + \frac{1}{4} = 2 + 0.25 = 2.25$

So, Haley uses 2.25 cup of sugar foe each pie.

Total cups of Sugar Haley has = 3 3/8 cups.

Now, $3 \frac{3}{8} = 3 + \frac{3}{8} = 3 + 0.375 = 3.375$

So, the total amount of sugar Haley has is 3.375 cups.

Now, $\textrm{Number of pies she can make} = \frac{\textrm{Total cups of sugar she has}}{\textrm{Cups of sugar required for each cake}}$

= $\frac{3.375}{2.25} =1.5$

or the number of pies she can make = 1.5

Hence, with 3 3/8 cups of sugar Haley can Make 1 1/2 pie.

8 0

3 years ago

3(n+1)= 5.4 enter the answer in the box n=?

NARA [144]

N= 0.8
this is the answer to ur problem

5 0

3 years ago

What change do you have to make to the graph of f (x) = 7x in order to graph the function g (x) = 7x+10?

cricket20 [7]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)= A( Bx+ C)+ D
\\\\
~~~~y= A( Bx+ C)+ D
\\\\
f(x)= A\sqrt{ Bx+ C}+ D
\\\\
f(x)= A(\mathbb{R})^{ Bx+ C}+ D
\\\\
f(x)= A sin\left( B x+ C \right)+ D
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\
\bullet \textit{ flips it upside-down if } A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if } B\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{ C}{ B}\\
~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by } D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{ B}$

now, with that template in mind, let's see

$\bf \stackrel{parent}{f(x)=7x}\qquad \qquad \qquad \stackrel{transformed}{g(x)=7x\stackrel{D}{+10}}$

D = 10, upwards shift of 10 units.

4 0

4 years ago