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Lorico [155]
3 years ago
14

A square is shown below. Which expression can be used to find the area, in square units, of the shaded triangle in the square?

Fraction 1 over 4 • 6•6 Fraction 1 over 2 •3•3
Fraction 1 over 4 •3•3
Fraction 1 over 2 •6•6

Mathematics
2 answers:
ELEN [110]3 years ago
7 0

Answer:

Area = \frac{1}{2} (6)(6)

Step-by-step explanation:

Area of Triangle = \frac{1}{2} (Base)(Height)

Where Base = 6 , Height = 6

Area = \frac{1}{2} (6)(6)

Area = 3 * 6

<u><em>Area = 18 units^2</em></u>

kotegsom [21]3 years ago
6 0

Answer:

Hey there!

The area of the shaded triangle covers exactly half the square.

Thus, first we find the area of the square, which is 6x6, or 36.

Finally, half of that is 1/4 times 6x6.(Fraction 1 over 2 •6•6 is correct)

Let me know if you need more help :)

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sattari [20]

Answer:

8.9 in.

Step-by-step explanation:

Area of triangle = (1/2) b. h

40 = (1/2) x*x

80 = x^2

square root (80) = x

x=> 8.944.... inch

Rounding off you get 8.9 in.

7 0
3 years ago
Ax + bx - 15 = 0;x<br> How do you solve this?
Pani-rosa [81]

Answer:

Step-by-step explanation:

ax+bx-15=0\\=> (a+b)x-15=0\\=>(a+b)x=15\\=>x=\frac{15}{(a+b)}

7 0
3 years ago
Help me it is below
Fofino [41]
I hope this helps you

7 0
3 years ago
Read 2 more answers
I added a screenshot of my question
Anastasy [175]
<h2>Answer:</h2>

y=\frac{1}{3}x+4

<h2>Explanation:</h2>

Slope-intercept form:

What the variables mean:

Y=the y axis

m=the slope

x=the x-axis

b=the y intercept (the point on the line that crosses the y-axis)

To Calculate the Slope:

1. Find any 2 points on the line

2. The number of units the second point is above the first is the numerator

3. The number of units the second point is to the right of the first is a denominator

(Please see the picture attached)

y=\frac{1}{3}x+b

To Find the Y-Intercept:

1. Find the point on the line that crosses the y-axis

(Please see the picture attached)

y=\frac{1}{3}x+4

5 0
3 years ago
Integrate ​G(x,y,z)equalsz over the parabolic cylinder yequalszsquared​, 0less than or equalsxless than or equals2​, 0less than
yKpoI14uk [10]

I gather you're supposed to compute the integral of G(x,y,z)=z over a surface S that is the part of the parabolic cylinder y=z^2 with 0\le x\le2 and 0\le z\le\frac{\sqrt{15}}2.

We can parameterize S by

\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k

with the given constraints on x and z. Take the normal vector to S to be

\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k

so that the surface element is

\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz

Then in the integral, we have

\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}

5 0
3 years ago
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