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3 years ago

If DB bisects < ADC and m<1=35° then m<2 is

Mathematics

2 answers:

svetlana [45]3 years ago

5 0

Angle 2 is also 35 degrees. Angle bisectors divide an angle into 2 congruent parts.

Finger [1]3 years ago

5 0

I'm not really sure guys I'm sorry I'm new here

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The 7th grade had a book drive. They donated 2/3 of the books to the st James school. Then donated 1/4 of the remaining books to

sergiy2304 [10]

Answer: 192 books

Step-by-step explanation:

2/3 + 1/4 = 11/12

16 remaining books = 1/12

16 x 11 = 176

176 + 16=192

7 0

2 years ago

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Graph the system of equations. then determine wheather the system has no solution, one solution, or infinitely many solutions. I

dimaraw [331]

Answer:

See below in bold.

Step-by-step explanation:

If we add the 2 equations we eliminate x and we get 2y =2.

So y = 1.

Substituting y = 1 in the second equation 1 = x - 3.

So x = 4.

A. One solution: (4, 1).

If we drew a graph we would have 2 lines which intersect at the point (4, 1).

8 0

3 years ago

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Does anybody have any tips for calculating slopes in Algebra 1???

I am Lyosha [343]

Answer:

The equations are really hard, so if you can find a slope interpreter or something that might be helpful.

I find it easier to just count the up and across.

Step-by-step explanation:

6 0

2 years ago

In a lab, a scientist is growing bacteria for a study. The scientist begins with a bacteria

Diano4ka-milaya [45]

Answer:

The answer is below

Step-by-step explanation:

The growth of the bacteria is in the form of an exponential growth. It is given by the formula:

$P(t)=ae^{rt}\\\\where\ t\ is\ the\ number\ of \ hours, P(t)\ is\ the \ population\ at\ t\ hours\\\and\ a=population\ at\ start$

At 2 hours, the population is 62 cells, hence:

$P(2)=ae^{2r}\\\\62=ae^{2r}\ \ .\ \ .\ \ .\ (1)$

After another 2 hours (4 hours), the population is 1 million:

$P(4)=ae^{4r}\\\\1000000=ae^{4r}\ \ .\ \ .\ \ .\ (2)\\\\Divide \ equation\ 2\ by\ equation\ 1:\\\\\frac{1000000}{62}=\frac{ae^{4r}}{ae^{2r}} \\\\16129=e^{2r}\\\\ln(e^{2r})=ln(16129)\\\\2r=9.688\\\\r=4.844$

Put r = 4.844 in equation 1

$62=ae^{2*4.844}\\\\62=16129a\\\\a=0.003844$

$P(t)=0.003844e^{4.844t}\\\\at \ start,t=0\\\\P(0)=0.003844$

4 0

3 years ago

A cone has a radius of 6.2 mm and a height of 10.8 mm. What is the volume of the cone to the nearest tenth? use π = 3.14

guapka [62]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the volumes.

Here given as, r= 6.2 mm, h = 10.8 mm, π=3.14

hence, the volume of the cone is

Volume = 1/3(πr^2h)

===> vol = 1/3(3.14*(6.2)^2*(10.8))

==> Vol = 1/3*(1303.57)

==> Vol = 434.52 mm^3

Hence the volume of the cone is 434.52 mm^3

7 0

2 years ago

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