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3 years ago

7

What is the lateral area of the square pyramid, to the nearest whole number?

1 answer:

FinnZ [79.3K]3 years ago

8 0

Answer:

351 m^2

btw

#1= 448 cm^2

#2=209pi cm^2

#3=351 m^2

and that's all I know hope it really helps we all need it sometimes (;

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One x-intercept for a parabola is at the point

Fudgin [204]

Answer:

(2,0) was already given so (-1,0) is the other one.

Step-by-step explanation:

So we are asked to use the quadratic formula.

To find the x-intercepts (if they exist) is use:

$\text{ If } y=ax^2+bx+c \text{ then the } x-\text{intercepts are } (\frac{-b \pm \sqrt{b^2-4ac}}{2a},0)$ .

Let's start:

Compare the following equations to determine the values for $a,b, \text{ and }c$ :

$y=ax^2+bx+c$

$y=4x^2-4x-8$

So

$a=4$

$b=-4$

$c=-8$

We are now ready to enter into our formula:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{4 \pm \sqrt{(-4)^2-4(4)(-8)}}{2(4)}$

$x=\frac{4 \pm \sqrt{16+16(8)}}{8}$

$x=\frac{4 \pm \sqrt{16(1+8)}}{8}$

$x=\frac{4 \pm \sqrt{16}\sqrt{1+8}}{8}$

$x=\frac{4 \pm 4\sqrt{9}}{8}$

$x=\frac{ 4 \pm 4(3)}{8}$

$x=\frac{4 \pm 12}{8}$

$x=\frac{4(1\pm 3)}{8}$

$x=\frac{1(1\pm 3)}{2}$

$x=\frac{1 \pm 3}{2}$

$x=\frac{1+3}{2} \text{ or } \frac{1-3}{2}$

$x=\frac{4}{2} \text{ or } \frac{-2}{2}$

$x=2 \text{ or } -1$

So the x-intercepts are (2,0) and (-1,0).

(2,0) was already given so (-1,0) is the other one.

7 0

3 years ago

the length of a large suv is 18 9/10, and the length of a small suv is 15 1/8 How much longer is the larger suv

puteri [66]

I think the answer is 14 13/40

8 0

2 years ago

Read 2 more answers

Which terms could be used as the first term of the expression below to create a polynomial written in standard form? Select five

vovikov84 [41]

Option C: $-6rs^5$

The last two terms given of the standard polynomial are:

$\cdots + 8r^2s^4 + 3r^3s^3$

This shows 2 patterns to us:

r 's degree is increasing per term in the given polynomial.

s 's degree is decreasing per term in the given polynomial.

The sum of powers of variates r and s is 2+4 = 3+3 = 6.

And since standard polynomial, the same pattern must be present in all terms of the series.

All options except C are not valid.

Option C has term $-6rs^5$ for which r's degree plus s's degree is 1 + 5 = 6.

And r's degree 1 < s's degree 5.

Thus its a valid term to be putted as first term in standard form of polynomial in consideration.

Thus Option C: $-6rs^5$ is the needed expression.

Learn more here:

brainly.com/question/10127580

4 0

2 years ago

R (-3,1) and S (-1,3) are points on a circle. If RS is a diameter, find the equation of the circle.

ELEN [110]

Answer:

$\sf (x+2)^2+(y-2)^2=2$

Step-by-step explanation:

If RS is the diameter of the circle, then the midpoint of RS will be the center of the circle.

$\sf midpoint=\left(\dfrac{x_s-x_r}{2}+x_r,\dfrac{y_s-y_r}{2}+y_r \right)$

$\sf =\left(\dfrac{-1-(-3)}{2}+(-3),\dfrac{3-1}{2}+1 \right)$

$\sf =(-2, 2)$

Equation of a circle: $\sf (x-h)^2+(y-k)^2=r^2$

(where (h, k) is the center and r is the radius)

Substituting found center (-2, 2) into the equation of a circle:

$\sf \implies (x-(-2))^2+(y-2)^2=r^2$

$\sf \implies (x+2)^2+(y-2)^2=r^2$

To find $\sf r^2$ , simply substitute one of the points into the equation and solve:

$\sf \implies (-3+2)^2+(1-2)^2=r^2$

$\sf \implies 1+1=r^2$

$\sf \implies r^2=2$

Therefore, the equation of the circle is:

$\sf (x+2)^2+(y-2)^2=2$

4 0

2 years ago

Determine if the given pair of lines is parallel, perpendicular or neither.

lana66690 [7]

It’s neither, and this is because the slope is not opposite or the same, the same y intercept yes...but there isn’t any parallel or perpendicular features...you can see this when you graph it :)

6 0

2 years ago