What is 7(5)-4x=51? Please

Mathematics

2 answers:

DaniilM [7]3 years ago

4 0

-4 = x
hope i helped ;)

podryga [215]3 years ago

3 0

7(5)-4x=51
7 times 5 is 35, so:
35-4x=51
add 4x to both sides:
35=4x+51
subtract 51 from both sides to get:
-16=4x
divide by 4 on both sides to get
-4=x

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2. Your friend wants to write the equation of a quadratic function 1 5. 3.r = 1) that has zeros at 3 and 3. They want to use the

Feliz [49]

Let's expand both options, your friend's and yours, as shown below

$\begin{gathered} y=(3x+1)(3x-5)=9x^2-12x-5=9(x^2-\frac{4}{3}x-\frac{5}{9}) \\ \text{and} \\ y=(x+\frac{1}{3})(x-\frac{5}{3})=x^2-\frac{4}{3}-\frac{5}{9} \end{gathered}$

Then, both equations are the same besides a constant that will not affect the zeros of the functions, as shown below

$\begin{gathered} y=(3x+1)(3x-5) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-1+1)(-1-5)=0 \\ \text{and} \\ x=\frac{5}{3} \\ \Rightarrow y=(5+1)(5-5)=0 \\ \end{gathered}$

And

$\begin{gathered} y=(x-\frac{5}{3})(x+\frac{1}{3}) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-\frac{1}{3}-\frac{5}{3})\cdot0=0 \\ x=\frac{5}{3} \\ \Rightarrow y=0\cdot(\frac{6}{3})=0 \end{gathered}$

Both your friend and you are correct. The functions are the same with exception of a constant that multiplies the whole function (a scale factor); despite that, the zeros are the same for both functions