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3 years ago

Determine which equation below can be solved to find the value of ∠A.

Mathematics

1 answer:

tia_tia [17]3 years ago

4 0

Answer:

Step-by-step explanation:

<A = 180 - (63+90)

<A = 180 - 63 - 90

<A = 180 - (63+45+<B )

<A = 180 -63 -45 -<B

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H6) A straight trail with aNIEuniform inclination of 15°leads from a lodge at an elevation of1000 feet to a mountain lake at ane

yaroslaw [1]

see the image to better understand the problem

we have that

sin(15)=6800/L -------> by opposite side divided by the hypotenuse

solve for L

L=6800/sin(15)

6 0

1 year ago

Math expressions that equal 13

qaws [65]

Step-by-step explanation:

There can be a lot of math expressions that may equal 13. Here are some of the expressions that equal 13.

9 + 4 = 13

15 - 2 = 13

1 × 13 = 13

Now, let us consider some interesting expression

$x^3-x^2+x+7$

If the value of x = 2, then the expression will be equal to 13.

i.e.

$2^3-2^2+2+7=13$ ∵ when x = 2

Considering a word problem

<em>'Tom has 10 oranges while Tony has 3 more oranges than Tom. How many oranges Tony has?</em>

Let 'x' be the number of oranges Tom has.

The number of oranges Tony has = x + 3

as Tom has 10 oranges, so x = 10

Therefore, the number of oranges Tony has = x + 3

= 10 + 3 = 13

Therefore, the number of oranges Tony has = 13

7 0

3 years ago

Lisa is buying a computer for $1,500. The sales tax rate is 8.5%. Which expression should be used to calculate t, the total cost

Sveta_85 [38]

Answer:

Step-by-step explanation:

Step one:

given data

price of computer= $1500

sales tax= 8.5%

Step two

Let us compute the amount in tax

tax= 8.5/100*1500

=0.085*1500

=$127.5

Let the total cost be t

hence

t=1500+0.085*1500

t=1500+127.5

t=$1627.5

6 0

3 years ago

5,000+600+7 in standard form (Sorry I forgot-)

goblinko [34]

Answer:

It is 5,607

Step-by-step explanation:

Its ok I forget slot to

7 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago