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3 years ago

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The graph of y=54x+1 is shown below. The coordinate (0,y) is a solution of the equation. Given the x–value of 0, what is the val

ue of the y–coordinate for the coordinate (0,y)

1 answer:

Serggg [28]3 years ago

7 0

the y value is one ? just insert the 0 into the x which would make the 54 disappear and leave you with y=1

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I need help with these 2 questions please help me with scie

Andre45 [30]

He's using the graduated cylinder method. I think the volume might be between 15 and 20. (It's kind of not obvious...)

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2 years ago

If a watermelon weighs 6 255 grams and a scales measures the weight 6 475 grams. What is the scales percent error

Liono4ka [1.6K]

Answer:scales percent error=3.51%

Step-by-step explanation:

P ercent error = Experimental value - Theoretical value /Theoretical value x 100

Theoretical value = 6255 grams

Experimental value ( use of scale ) = 6475grams

Percent error =(6475- 6255)grams / 6255grams x 100

=220/6255 x 100

=3.51%

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3 years ago

Find the common ratio of the geometric sequence –5,–30, -180,...

Murljashka [212]

Answer:

6

Step-by-step explanation:

You have to find what to multiply by to get to the next term

-30 ÷ -5 = 6

- 180 ÷ -30 = 6

therfore the common ratio is 6

5 0

2 years ago

Read 2 more answers

I’m struggling with this question need some help

Sonbull [250]

Since the volume of a rectangular prism is V=lwh, we can isolate for h to obtain that h=V/(lw).

So, for the first prism, the height is 117/(4*6.5) = 4.5

Similarly, for the second prism, the height is 175.5/(6.5 * 4) = 6.75

So, the difference in heights is 6.75-4.5=2.25 = 2 1/4 inches

8 0

1 year ago

Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0

solong [7]

Answer:

The solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3$

Let us determine all possible solutions for f(x) = 0

$0=4x^2-4x+3$

switch both sides

$4x^2-4x+3=0$

subtract 3 from both sides

$4x^2-4x+3-3=0-3$

simplify

$4x^2-4x=-3$

Divide both sides by 4

$\frac{4x^2-4x}{4}=\frac{-3}{4}$

$x^2-x=-\frac{3}{4}$

Add (-1/2)² to both sides

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2$

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{1}{2}=\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}$ ∵ $\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}$

as

$\sqrt{-1}=i$

so

$x-\frac{1}{2}=i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

also solving

$x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Therefore, the solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

4 0

2 years ago