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3 years ago

Use the following information for Questions 1–3:

Mathematics

1 answer:

UkoKoshka [18]3 years ago

6 0

It is easier to do one question at a time

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Find the surface area of the present.<br><br> square : 15cm width and 15cm length

VladimirAG [237]

Answer:

The area of the square is 225 cm².

Step-by-step explanation:

Since the geometric figure has a shape of a square, which is a special form of rectangle that has all the sides with the same length, then it's surface area is given by the following formula:

area = length*width = length^2

We can solve for the area by applying the data from the problem, we have:

area = (15)^2

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3 years ago

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3 years ago

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Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)

sesenic [268]

Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3> or , $u = -5i-4j , v=-4i-3j$

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

$u.v =|u|(|v|)cosx$ , where x is angle between u & v !

⇒ $u.v =|u|(|v|)cosx$

⇒ $cosx =\frac{u.v}{|u|(|v|)}$

Now , $u.v = (-5i-4j)(-4i-3j)$

⇒ $u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)$

⇒ $u.v = 20+12$ { $i(j) = j(i) =0$ }

⇒ $u.v = 32$

Now , Modulus of any vector $r = xi+yj$ is $|r| = \sqrt{x^{2}+y^{2}}$ So ,

$|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

Putting all these values in equation $cosx =\frac{u.v}{|u|(|v|)}$ we get:

⇒ $cosx =\frac{32}{5(\sqrt{41})}$

⇒ $cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})$

⇒ $x =cos^{-1}(1)$ { $cos0 = 1$ }

⇒ $x =0$ °

Therefore , Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

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3 years ago