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3 years ago

Could someone confirm if I was correct?

Mathematics

1 answer:

mart [117]3 years ago

3 0

compare to the correct answer on the picture. If not helped you can use a graph calculator and it should let you know wether or not you’re correct

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Answer:

Step-by-step explanation:

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3 years ago

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Find ab. round to the nearest tenth if necessary

Brrunno [24]

What is the equation?

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3 years ago

A triangle has vertices at (-1,5), (4,2), and (0,0). What is the perimeter of the triangle

yan [13]

The perimeter of the triangle is 15.4 units.

Step-by-step explanation:

The triangle has vertices at (-1,5), (4,2), and (0,0).

To find the perimeter of the triangle, you need to find the distance of the three points which is the length of the three sides of the triangle.

Let us consider the points as A (-1,5) and B (4,2) and C (0,0).

The distance formula is given by :

⇒ $\sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$

Distance of AB :

A (-1,5) ⇒ (x1,y1)

B (4,2) ⇒ (x2,y2)

⇒ $\sqrt{(4+1)^{2}+(2-5)^{2} }$

⇒ $\sqrt{5^{2}+-3^{2} }$

⇒ $\sqrt{25+9}$

⇒ $\sqrt{34}$

⇒ 5.8

∴ The length of the side AB of the triangle is 5.8 units.

Distance of BC :

B (4,2) ⇒ (x1,y1)

C (0,0) ⇒ (x2,y2)

⇒ $\sqrt{(0-4)^{2}+(0-2)^{2} }$

⇒ $\sqrt{16+4 }$

⇒ $\sqrt{20}$

⇒ 4.5

∴ The length of the side BC of the triangle is 4.5 units.

Distance of CA :

C (0,0) ⇒ (x1,y1)

A (-1,5) ⇒ (x2,y2)

⇒ $\sqrt{(-1-0)^{2}+(5-0)^{2} }$

⇒ $\sqrt{1+25}$

⇒ $\sqrt{26}$

⇒ 5.1

∴ The length of the side CA of the triangle is 5.1 units.

Now, the perimeter of the triangle is given by :

Perimeter = Sum of the lengths of all three sides of the triangle.

⇒ 5.8 + 4.5 + 5.1

⇒ 15.4 units.

∴ The perimeter of the triangle is 15.4 units.

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3 years ago

Polygon ABCDE is reflected across the x-axis to form polygon A′B′C′D′E′. Then polygon A′B′C′D′E′ is dilated by a scale factor of

elena-14-01-66 [18.8K]

It's dilated by 0.5 not 2.

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3 years ago

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Lana71 [14]

The sub-boxes will have dimensions $\frac{2-0}{2} \times \frac{2-0}{2} \times \frac{2-0}{2} =1\times1\times1=1 \ cubic \ units$

x sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $x= \frac{1}{2}$ and $x= \frac{3}{4}$
y sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $y= \frac{1}{2}$ and $y= \frac{3}{4}$
z sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $z= \frac{1}{2}$ and $z= \frac{3}{4}$ 

Let $f(x,y,z)=\cos{(xyz)}$

$\int\limits \int\limits \int\limits {f(x,y,z)} \, dV \approx f\left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{3}{4} \right)$
$+f\left( \frac{3}{4} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{3}{4} \right) \\ \\ \approx\cos{ \frac{1}{8} }+\cos{ \frac{3}{16} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{9}{32} }+\cos{ \frac{27}{64} } \\ \\ \approx0.9922+0.9825+0.9825+0.9607+0.9825+0.9607+0.9607 \\ +0.9123 \\ \\ \approx\bold{7.734}$

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3 years ago