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Lemur [1.5K]
3 years ago
5

What is the length (magnitude) of vector (6,-3)? Sqrt 53 Sqrt 2 3 3 sqrt 5

Mathematics
2 answers:
vekshin13 years ago
7 0
Hope this would help you

zysi [14]3 years ago
7 0

Answer:

(D)

Step-by-step explanation:

The given vector is  (6,-3), the magnitude of the given vector is gievn as:

m=\sqrt{(x)^2+(y)^2}

Putting x=6 and y=-3, we get

m=\sqrt{(6)^2+(-3)^2}

m=\sqrt{36+9}

m=\sqrt{45}

m=3\sqrt5

Thus, the length of the given vector is 3\sqrt5 units.

Therefore, option D is correct.

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Match each value with its formula for ABC.
Ivahew [28]

For the triangle ABC use the sine theorem:

\dfrac{a}{\sin A}= \dfrac{b}{\sin B}= \dfrac{c}{\sin C}.

1. From \dfrac{a}{\sin A}= \dfrac{b}{\sin B} you have \dfrac{a}{b} \cdot \sin B=\sin A.

2. From \dfrac{b}{\sin B}= \dfrac{c}{\sin C} you have \dfrac{b}{c} \cdot \sin C=\sin B.

3. From \dfrac{a}{\sin A}= \dfrac{c}{\sin C} you have \dfrac{c}{a} \cdot \sin A=\sin C.

4. From \dfrac{a}{\sin A}= \dfrac{c}{\sin C} you have \dfrac{\sin A}{\sin C} \cdot c=a.

5. From \dfrac{a}{\sin A}= \dfrac{b}{\sin B} you have \dfrac{\sin B}{\sin A} \cdot a=b.

6. From \dfrac{b}{\sin B}= \dfrac{c}{\sin C} you have \dfrac{\sin C}{\sin B} \cdot b=c.

5 0
4 years ago
Read 2 more answers
What is the first step when rewriting y=3x^2+9x-18
Pachacha [2.7K]

Answer:

do the pemdas

p=()

e=X^x

md= multiply then devide

as= add then subtract

Step-by-step explanation:

y=3x^2+9x-18

y=9x+9x-18

18x-18=y

3 0
3 years ago
1000 households were surveyed. 275 households own a desktop computer, 455 households own a DVD player, 405 households own two ca
icang [17]

Answer:

Step-by-step explanation:

From the given information,

Suppose

X represents the Desktop computer

Y represents the DVD Player

Z represents the Two Cars

Given that:

n(X)=275

n(Y)=455

n(Z)=405

n(XUY)=145

n(YUZ)=195

n(XUZ)=110

n((XUYUZ))=265

n(X ∩ Y ∩ Z) = 1000-265

n(X ∩ Y ∩ Z) = 735

n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)

145 = 275+455 - n(X ∩ Y)

n(X ∩ Y) = 585

n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)

195 = 455+405-n(Y ∩ Z)

n(Y ∩ Z) = 665

n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)

110 = 275+405-n(X ∩ Z)

n(X ∩ Z) = 570

a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)

n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735

n(X ∪ Y ∪ Z) = 50

c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)

n(X ∪ Y ∪ C') = 145-50

n(X ∪ Y ∪ C') = 95

6 0
3 years ago
All you need is in the photo <br><br>please answer step by step ​
sasho [114]
Carol, download a app called connects way better
7 0
3 years ago
PLSS HELP ILL GIVE YOU BRAINLIEST!!
uysha [10]

Answer:

Step-by-step explanation:

Q1)

Use Phythogoras theorem:

(AC)^2=(BC)^2+(AB)^2\\BC^2=AC^2-AB^2\\BC^2=6^2-4^2\\BC^2=36-16\\BC^2=20\\\\Apply\ square\ on\ both\ sides\\\\\sqrt{BC^2}=\sqrt{20}  \\BC=\sqrt{20} \\BC=\sqrt{10(2)} \\BC=\sqrt{5(2)(2)} \\BC=2\sqrt{5}

Q2)

Apply phythogoras theorem:

CD^2=CE^2+DE^2\\CD^2=7^2+9^2\\CD^2=49+81\\CD^2=130\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{CD^2} = \sqrt{130} \\\\CD=\sqrt{130} \\\\CD=11.4

Q3)

Apply phythogoras theorem again:

BC^2=BD^2+CD^2\\BC^2=7^2+13^2\\BC^2=49+169\\BC^2=218\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{BC^2} =\sqrt{218}  \\\\BC=\sqrt{218} \\\\BC=14.76

I have an attached an image for Question 2 for better understanding the length of DE in question equals 15 - 6 = 9

4 0
3 years ago
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