All categories

2 years ago

Please help Find the area in square units of ABC△ plotted below.

Mathematics

1 answer:

neonofarm [45]2 years ago

6 0

Answer:

Use the distance formula on both points AC and AB.

Distance formula is this:

\begin{gathered}d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\\\d=\sqrt{(1--5)^2+(8--7)^2} \\\\d=\sqrt{(6)^2+(15)^2} \\\\d=\sqrt{36+225} \\\\d=\sqrt{261} \\\\\end{gathered}d=(x2−x1)2+(y2−y1)2d=(1−−5)2+(8−−7)2d=(6)2+(15)2d=36+225d=261

Distance for AC is 16.16

Now do the same with the numbers for AB and get the distance of 5.39

2. To get the area, use the formula 1/2 x base x height

AB is the base and AC is the height.

1/2 x 16.16 x 5.39 = 43.55

the answer is 43.5

You might be interested in

Check the picture please if you're good at math!!

Jet001 [13]

Answer:

Step-by-step explanation:

The triangles aren't similar.

The triangles share a side so we cant use anything involving SSS SAS.

The triangles has only known congruent angle so we cant use AA.

5 0

3 years ago

Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than $A(t)=A_{0}e^{kt}$ . for example, if an initial population P₀ requires n units of time to triple, then the function $P(t)=P_{0}(3)^{\frac{t}{n} }$ models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form $P(t)=P_{0}(3)^{\frac{t}{n} }$ that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form $A(t)=A_{0}e^{kt}$ .

Answer: a) $P(t)=50(3)^{\frac{t}{30} }$

b) P(t) = 280 insects

c) t = 74 days

d) $A(t)=50e^{0.037t}$

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

$P(t)=50(3)^{\frac{t}{30} }$

b) In t = 47 days:

$P(t)=50(3)^{\frac{t}{30} }$

$P(47)=50(3)^{\frac{47}{30} }$

$P(47)=50(3)^{1.567}$

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

$750=50(3)^{\frac{t}{30} }$

$\frac{750}{50}=(3)^{\frac{t}{30} }$

$(3)^{\frac{t}{n} }=15$

Using the property Power Rule of logarithm:

$log(3)^{\frac{t}{30} }=log15$

$\frac{t}{30}log(3)=log15$

$t=\frac{log15}{log3} .30$

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

$A(t)=A_{0}e^{kt}$

$3A_{0}=A_{0}e^{30k}$

$3=e^{30k}$

Use Power Rule again:

$ln3=ln(e^{30k})$

$ln3=30k$

$k=\frac{ln3}{30}$

k = 0.037

Equation for exponential growth will be:

$A(t)=50e^{0.037t}$

3 0

3 years ago

Hello whats 2+2 :p thanks

Ugo [173]

Answer:

Step-by-step explanation:

lol ez points

8 0

3 years ago

Read 2 more answers

Will mark branlist for the first one to help

Nikitich [7]

Answer:

17°

Step-by-step explanation:

73°= EGB

EGB+ x° = 90°

so x° = 90°-73° =17°

8 0

3 years ago

Please help exponents and scientific notation

Anna007 [38]

Alright! When you have a constant to a power times another constant to a power (ex. [x^3 times x^3] ) you simply add the powers and keep the base [x^6]. When you have a power to a power (ex. [(12^3)^3] ) you multiply the powers and keep the base [12^9]. When you have a constant to a power divided by a constant to a power (ex. [ x^2 divided by x^5] ) you subtract the powers and keep the base. It's hard to see the questions, so I'll leave this here for you to use as a guide.

6 0

3 years ago