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3 years ago

Please guys I need your help for this question ASAPPp!!!

Mathematics

1 answer:

o-na [289]3 years ago

8 0

yea they're equivalent cause to solve the second equation sqrt of x+1 plus 3 equals 8 you must subtract 3 from both sides and when you do that you get sqrt of x+1=5 and thats the same as the first equation

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Thepotemich [5.8K]

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3 years ago

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Tasya [4]

Answer:

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Step-by-step explanation:

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3 years ago

Verify the identity cos quantity x plus pi divided by two = -sin x

In-s [12.5K]

Answer:

Identity is verified.

Step-by-step explanation:

We have to verify the identity $cos(x+\frac{\pi }{2}) = (- sinx)$

To prove any identity we always prove one side(either left hand side or right hand side) of the equation equal to the other side.

In this identity we take the left hand side first

$cos(x+\frac{\pi}{2})$

$=cosx\times cos(\pi/2)-sinx\times sin(\pi/2))$ (as we know cos(a+b) = cosa×cosb-sina×sinb)

$= cosx\times0-sinx\times1$

$= 0-sinx$

= - sinx ( Right hand side)

Hence identity is proved.

6 0

3 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

8 0

2 years ago