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3 years ago

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Mathematics

1 answer:

Jlenok [28]3 years ago

8 0

Answer:

Step-by-step explanation:

yay thats the first question i answered

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Please answer correctly !!!!!!!!!! Will mark brainliest !!!!!!!!!!!!

Leto [7]

Answer:

see explanation

Step-by-step explanation:

(1)

Given

g(r) = (r + 14)² - 49

To obtain the zeros, let g(r) = 0 , that is

(r + 14)² - 49 = 0 ( add 49 to both sides )

(r + 14)² = 49 ( take the square root of both sides )

r + 14 = ± $\sqrt{49}$ = ± 7 ( subtract 14 from both sides )

r = - 14 ± 7, then

r = - 14 - 7 = - 21 ← smaller r

r = - 14 + 7 = - 7 ← larger r

(2)

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

g(r) = (r + 14)² - 49 ← is in vertex form

with vertex = (- 14, - 49 )

4 0

3 years ago

Why is 1+(-5) equal to -4

erica [24]

1 + (-5) becomes 1 - 5 because you are adding a negative number.

1 - 5 is -4

5 0

4 years ago

the question is in the picture below, if you could show all your work so i can see how to do it, that would be appreciated.

ki77a [65]

m∠R = 27.03°

Solution:

Given In ΔQRP, p = 28 km, q = 17 km, r = 15 km

To find the measure of angle R:

Law of cosine formula for ΔQRP:

$r^{2}=p^{2}+q^{2}-2 pq \cos R$

Substitute the given values in the above formula.

$(15)^{2}=(28)^{2}+(17)^{2}-2 (28)(17) \cos R$

$225=784+289-952 \cos R$

$225=1073-952 \cos R$

Switch the given equation.

$1073-952 \cos R=225$

Subtract 1073 from both side of the equation.

$-952 \cos R=-848$

Divide by –952 on both sides.

$$ \cos R=\frac{106}{119}$

$$ R=\cos^{-1} \left(\frac{106}{119}\right)$

$R=27.03^\circ$

Hence m∠R = 27.03°.

7 0

4 years ago

Pls answer this i want to know ok ok ok ok ok

bogdanovich [222]

Step-by-step explanation:

$given \\ (x + \frac{1}{x} ) = 7 \\ now \\ (x + \frac{1}{x} ) ^{2} \\ (7) ^{2} (given) \\ 49 \\ again \\ {x}^{2} + ( \frac{1}{x} ) ^{2} \\ (x + \frac{1}{x}) ^{2} - 2 \times x \times \frac{1}{x} \\ (7) ^{2} - 2 \\ 49 - 2 \\ 47$

6 0

3 years ago

There are 527 pencils,646 erasers,748 sharpeners . these are to be put in separate packets containing the same no. of items. fin

Ksju [112]

$To \; find \; the \; maximum \; number \; of \; items \; possible \; in each \; packet,\\we \; need \; to \; find \; the \; Greatest \; Common \; Factor\\ \; of \; 527 \; Pencils, \; 646 \; erasers \; and \; 748 \; sharpeners.\\\\We \; need \; to \; list \; the \; Factors \; of \; each \; number \; as \; given \; below\\ \; and \; identify \; the \; greatest \; common \; factor.$

$The \; factors \; of \; 527 \; are:\\1, \; 17, \; 31, \; 527\\\\The \; factors \; of \; 646 \; are:\\1, \; 2, \; 17, \; 19, \; 34, \; 38, \; 323, \; 646\\\\The \; factors \; of \; 748 \; are:\\1, \; 2, \; 4, \; 11, \; 17, \; 22, \; 34, \; 44, \; 68, \; 187, \; 374, \; 748\\\\Then \; the \; greatest \; common \; factor \; is \; 17.$

$Conclusion: \\There \; will \; 31 \; Pencils, \; 38 \; Erasers,\\ \; and \; 44 \; Sharpeners \; in \; each \; of \; SEVENTEEN \; packet.$

4 0

3 years ago