What would this one be?

What is e-1/2f when e=15 and f=2

dusya [7]

Answer:

14

Step-by-step explanation:

Replace with the values

15-1/2(2) = 15-1 = 14

3 0

4 years ago

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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

Determine the most precise name for ABCD (parallelogram, rhombus, rectangle, or square). Explain how you determined your answer.

Sonja [21]

<h3>Answer: Rhombus</h3>

======================================================

Reason:

Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2, y_2) = (7,6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-7)^2 + (5-6)^2}\\\\d = \sqrt{(-4)^2 + (-1)^2}\\\\d = \sqrt{16 + 1}\\\\d = \sqrt{17}\\\\d \approx 4.1231\\\\$

Segment AB is exactly $\sqrt{17}$ units long, which is approximately 4.1231 units.

If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.

-------------------------

Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.

Let's find the slope of AB.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2,y_2) = (7,6)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{6 - 5}{7 - 3}\\\\m = \frac{1}{4}\\\\$

Segment AB has a slope of 1/4.

Do the same for BC

$B = (x_1,y_1) = (7,6) \text{ and } C = (x_2,y_2) = (6,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 6}{6 - 7}\\\\m = \frac{-4}{-1}\\\\m = 4\\\\$

Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.

We don't have perpendicular lines, so the interior angles are not 90 degrees each.

Therefore, this figure is not a rectangle and by extension it's not a square either.

The best description for this figure is a <u>rhombus</u>.

4 0

2 years ago

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sarah and her six friends are decorating picture frans with ribbon they have 5 rolls of ribbon to share evenly

Leya [2.2K]

Http://lbcsd.innersync.com/sites/jhayes/documents/U1M3Gr6pg.49-502015JHayes.pdf

5 0

3 years ago

H(x)=2x3 – 3<br> Find h(2)

djverab [1.8K]

When it tells you to find h(2), that means you plug in 2 for x.
If 2x^3 (to the third power)
2(2)^3-3

1. PEMDAS, exponents first
2^3=8

2. multiply
8x2=16

3. subtract
16-3=13

Hope this helps!

8 0

3 years ago