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2 years ago

Round 523,009.974 to the nearest tenth

Mathematics

1 answer:

geniusboy [140]2 years ago

4 0

Answer:

Step-by-step explanation:

523,010.0

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17. Find the ratio of the perimeter to the area of the square shown.<br> x + 3<br> x + 3

Serga [27]

Answer:

4x+12/x^2+9

Step-by-step explanation:

Perimeter

x+3+x+3+x+3+x+3

Area

(x+3)(x+3)

7 0

3 years ago

(Cosx)/(1+sinx) - (1+sinx)/(cos) solve

Rina8888 [55]

2/cos x

........................

8 0

3 years ago

Read 2 more answers

A 1st degree polynomial function whose graph intercepts the y-axis at 8.

asambeis [7]

Your answer can be anything in the form y = mx+8 where you replace m with any real number.

You start with y = mx+b, and then replace the b with the y intercept 8.

The y intercept is where the polynomial crosses the y axis.

The value of m does not matter. So you could have y = 2x+8 or y = 3x+8 for instance. Replace m with whatever your favorite number is.

4 0

3 years ago

James has $5.00 more than three times what Catherine has. Together they have $41.00. How much does each student have?

Tpy6a [65]

Hey.

acc'g to the conditions :

Let Catherine has x
and James has 5 + 3x

Then,

$5 + 4x = 41 \\ \\ \implies 4x +5 -5 = 41-5 \\ \\ \implies 4x = 36 \\ \\ \implies x= \dfrac{36}{4} \\ \\ \implies x = 9$

Hence

• Catherine has = x = $9,
• and James has = 5 +3x = 5 + 3(9) = $32

Thanks.

7 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago