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DedPeter [7]
2 years ago
9

Calculate the area of kite PQRS.

Mathematics
2 answers:
Sonbull [250]2 years ago
6 0
Check the picture below

so, just get the area of each triangle, and sum them up, that's the area of the kite

Bogdan [553]2 years ago
4 0
A=1/2[(15+23)*(8+8)]
   =1/2(608)
   =304
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EAST MATH PLS HELP ASAP PLEASE PLEASE
serious [3.7K]
4(x+7)=8x-20

4x+28=8x-20

28=4x-20

48=4x

x=12
7 0
3 years ago
Jason’s savings account has a balance of $2179. After 5 years , what will the amount of interest be at 6% compounded quarterly?
satela [25.4K]

Answer:

$755.80

Step-by-step explanation:

Determine the compound amount first and then subtract the principal from it, to find the amount of interest.

The compound amount formula is A = P (1 + r/n)^(nt), where

P is the initial principal, r is the interest rate as a decimal fraction, n is the number of compounding periods per year, and t is the number of years.  Here, P = $2179; t = 5 yrs; r = 0.06; and n = 4 (quarterly compounding).

We get:

A = $2179(1 + 0.06/4)^(4*5), or $2179(1.015)^20, or $2179(1.347) = $2937.80.

The compound amount is $2934.80.  Subtracting the $2179 principal results in the interest earned:  $755.80.

5 0
3 years ago
Find the distance between (3,4) and (4,-6) if necessary, round to the nearest tenth.
Ratling [72]

Answer:

The distance is:

d = 10.0 units (Rounded to the nearest the Tenths Place)

Step-by-step explanation:

Given the points

  • (3,4)
  • (4,-6)

The distance 'd' between (3,4) and (4,-6)

\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

substituting the points values

   =\sqrt{\left(4-3\right)^2+\left(-6-4\right)^2}

   =\sqrt{1+10^2}

   =\sqrt{1+100}

   =\sqrt{101}

   =10.0  units (Rounded to the nearest the Tenths Place)

Thus, the distance is:

d = 10.0 units (Rounded to the nearest the Tenths Place)

4 0
3 years ago
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
3 years ago
You stand 20 feet from a flagpole and sight the top at a 40 degree angle of elavatuon. How tall is the pole?
LenKa [72]
Tan 40 = h / 20  where h = height of the pole.

h = 20 tan 40 =  16.78 feet
8 0
3 years ago
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