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3 years ago

For this isosceles triangle, find the measure of angles h and m.

Mathematics

1 answer:

Genrish500 [490]3 years ago

7 0

Hello,

x + x + 128° = 180°

2x + 128° = 180°

2x = 180° - 128° = 52°

x = 52° / 2 = 26°

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Forty- two people attended s meeting about a zoning ordinance. Six more of the attendees voted in favor of the ordinance than th

kumpel [21]

Let the attendees in opposition be x. This means that the attendees in favour are x+6.
That means x+x+6=42, which is 2x+6=42, so 2x=36 and x=18. Therefore, there are 18 attendees who opposed it.

3 0

2 years ago

A² +6²<br> a-b<br> if a = 3 and b = 4<br> Evaluate each expression using the variable replacements.

Arturiano [62]

Knowing what the variables are means that all we have to do is substitute:
[ ( 3 ) ^2 + ( 4 ) ^2 ] / 4 - 3 =
( 9 + 16 ) / 1 =
25

3 0

2 years ago

Read 2 more answers

Which graph represents x - 1 ≤ 3

MAVERICK [17]

The first graph.

The equation is x - 1 $\leq$ 3
In order to solve it, you need to get x by itself.
Add 1 to both sides. 3+1=4
x $\leq$ 4

Since this is your sign $\leq$ the arrow has to point towards the left, and the circle needs to be colored in.

3 0

3 years ago

I ned help on homework pls help

zmey [24]

Answer:

Step-by-step explanation:

4 0

2 years ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$

d) $t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)$

$N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h$

6 0

3 years ago