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myrzilka [38]
3 years ago
8

What is the value of c such that

lt="x^{2}+14x+c" align="absmiddle" class="latex-formula"> is a perfect-square trinomial?
Mathematics
2 answers:
Alex73 [517]3 years ago
5 0
Your answer will be 49
suter [353]3 years ago
3 0

Answer:

49

Step-by-step explanation:

To find the 'c' value c=(b/2) to the second power. divide the coefficient of x by 2 and square the result.

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Angie’s Bake Shop makes birthday chocolate chip cookies that cost $2 each. Angie expects that 10% of the cookies will crack and
wlad13 [49]

Answer:

Step-by-step explanation:

the value of the cookies can be 1.20

8 0
3 years ago
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Marco read that a honey bee can fly up to 2.548meyers per second. he rounded the number to 2.55. to witch place valuedid marcoro
Kay [80]
He rounded to the hundredths place.
4 0
4 years ago
Please can someone answer this.
ivanzaharov [21]

Answer:

  •  x = -4, x = 0, x = 1

Step-by-step explanation:

x is a factor of all terms, so x=0 is a zero. (Eliminates choices 1 and 5.)

The sum of coefficients is 0, so x=1 is a zero. (Eliminates choices 3 and 4.)

Reversing the sign of the odd-degree terms gives signs of -++, so there is one sign change, hence one negative real root (by Descartes' rule of signs). This confirms choice 2 as the answer.

___

Of course, your graphing calculator can answer this almost as quickly.

7 0
3 years ago
The sum of two consecutive numbers is 71. Find the smallest number in the list
bazaltina [42]
35+36=71
35 is the answer
5 0
4 years ago
When n=343 college students are randomly selected and surveyed, it is found that x=110 own a car. Find a 99% confidence interval
Liono4ka [1.6K]

Answer:

The margin of error will be "0.65". A further explanation is provided below.

Step-by-step explanation:

The given values are:

n = 343

x = 110

At 99% confidence level,

\alpha = 1-99%

  =1-0.99

  =0.01

then,

\frac{\alpha}{2} =\frac{0.01}{2}

  =0.005

or,

Z_{\frac{\alpha}{2} }=Z_{0.005}

     =2.576

Now,

The point estimate will be:

⇒  \hat{P}=\frac{x}{n}

⇒      =\frac{110}{343}

⇒      =0.321

or,

⇒  1-\hat{P}=1-0.321

⇒            =0.679

The margin of error will be:

⇒  E=Z_{\frac{\alpha}{2} }\times \sqrt (\frac{(\hat{P}\times (1 - \hat{P})) }{n} )

On substituting the above values, we get

⇒      =2.576\times \sqrt{\frac{0.321\times 0.679}{343} }

⇒      =2.576\times \sqrt{\frac{0.217959}{343} }

⇒      =0.065

hence,

⇒  \hat{P}-E

⇒  0.321-0.065

⇒  0.256,0.386

6 0
3 years ago
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