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2 years ago

What are the solution(s) to the quadratic equation 50 – x2 = 0?

Mathematics

2 answers:

Rufina [12.5K]2 years ago

8 0

Answer:

x=5√2 or x=−5√2

Step-by-step explanation:

Let's solve your equation step-by-step.

50−x^2=0

Step 1: Simplify both sides of the equation.

−x^2+50=0

Step 2: Subtract 50 from both sides.

−x^2+50−50=0−50

−x^2=−50

Step 3: Divide both sides by -1.

−x^2 −1 = −50 −1

x^2=50

Step 4: Take square root.

x=±√50

x=5√2 or x=−5√2

Answer:

x=5√2 or x=−5√2

MARK ME BRAINLIEST PLEASE

Talja [164]2 years ago

6 0

Answer:

Step-by-step explanation:

took on ed

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In a class of 32 students, the mean height of the 14 boys is 1.56 metres. The mean height of all 32 students is 1.515 metres.

Mandarinka [93]

Answer:

1.48 m

Step-by-step explanation:

mean height of 14 boys is 1.56 m, so their total height is 14 * 1.56 = 21.84 m

mean height of 32 students is 1.515 m, so their total height is 48.48

48.48 - 21.84 = 26.64 (total height of the girls)

mean height of the girls is 26.64/18 = 1.48 m

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DedPeter [7]

y= -1/2x-3

y= -7/3x-5

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I hope that this helps

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2 years ago

Pencils come in boxes of 10. How many boxes should Erika buy if she needs 127 pencils?

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2 years ago

If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

Ilya [14]

This is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the "common difference" "d".

now, we know the 17th term is 34, let's go to the 15term then

34 + d
34 + d + d
34 + d + d + d
34 + d + d + d + d
34 + d + d + d + d + d
34 + d + d + d + d + d + d
34 + d + d + d + d + d + d + d
34 + d + d + d + d + d + d + d + d

now, notice, we got to the 15th term, which is 34 + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

$\bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}$

ok, now that we know what the common difference is, let's find the first term in the sequence using the 7th term of 34,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=5\\
a_7=34
\end{cases}
\\\\\\
a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5
\\\\\\
34=a_1+30\implies \boxed{4=a_1}$

and now let's use those 2 found folks, to get the 21st term,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=4\\
d=5\\
n=21
\end{cases}
\\\\\\
a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104$

3 0

2 years ago

A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for t

hoa [83]

Answer:

Since p value <0.1 accept the claim that oven I repair costs are more

Step-by-step explanation:

The data given for two types of ovens are summarised below:

Group Group One Group Two

Mean 85.7900 78.6700

SD 15.1300 17.8400

SEM 1.9533 2.3840

N 60 56

Alpha = 10%

$H_0: \mu_1 - \mu_2 =0\\H_a: \mu_1 - \mu_2> 0$

(Right tailed test)

The mean of Group One minus Group Two equals 7.1200

df = 114

standard error of difference = 3.065

t = 2.3234

p value = 0.0219

If p value <0.10 reject null hypothesis

4) Since p value <0.1 accept the claim that oven I repair costs are more

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2 years ago