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Karo-lina-s [1.5K]

3 years ago

7

HEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

PPPPPPPPP jk have some points

2 answers:

d1i1m1o1n [39]3 years ago

6 0

Step-by-step explanation:

Yeah thanks really much man, I appreciate it

mr_godi [17]3 years ago

5 0

Answer:

Thank you!

Step-by-step explanation:

Have a great day and stay safe :)

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)) Omar used 28 centimeters of tape to wrap 4 presents. How much tape will Omar need in

pav-90 [236]

Answer:

He would need 42 cm *or 14cm

Step-by-step explanation:

28/4 is 7

that means that he used 7 cm of tape per present

6 x 7 is 42

that means that he would need 42 cm of tape

4 to 28

6 to 42

*If he already has 4 presents done and he needs to 2 more present then he would need 14 cm, 2 x 7 is 14

8 0

3 years ago

Vera_Pavlovna [14]

Answer:

arc AC = 111°

Step-by-step explanation:

∠APC is a central angle, so arc AC is equal to the measure of ∠APC.

∠APC and ∠BPC are supplementary. So, m∠APC + m∠BPC = 180

m∠APC + 69 = 180

m∠APC = 111

arc AC = 111°

8 0

3 years ago

Which of the following is equivalent to log34

olasank [31]

hope it helps u...

plz mark as brainliest...

4 0

4 years ago

Convert to an improper fraction. Type in your answer with the negative in the numerator. -8 3/4 =

Sergeu [11.5K]

Answer:

-32/4

Step-by-step explanation:

-8 times 4= -32 put -32 over 4

8 0

3 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

3 years ago

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