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N76 [4]
3 years ago
5

Find the area, help please

Mathematics
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

Area- 22.5

Step-by-step explanation:

To find the area in a triangle you will want to multiple the base with the height then divide it by two. in this photo you have two options 9X5/2 or 7.5X6/2

remember- a= 1/2 b X h

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the ratio of the angle measures of a triangle are 3:3:4 what angle measures the triangles classification ​
TEA [102]

Step-by-step explanation:

sum of the ratio=3+3+4=10

now,

angle I = 3/10×180=54

angle II= 3/10×180=54

angle III=4/10×180=72

that shows th given triangle is isoceles

6 0
2 years ago
What is the equation for (0,3) and (6,6)
Alexxandr [17]
The answer is y= — 1/2x + 3
6 0
2 years ago
Use the elimination method to solve the system of equations.<br><br>3x-5y= -16<br><br>2x+5y=31
Sholpan [36]
5x = 15
x = 3

3(3) - 5y = -16
9 - 5y = -16
5y = -16 - 9
5y = - 25
y = - 5

(3, -5)
8 0
3 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
How can we solve proportions? ( 2 words)
Vilka [71]

Answer:

Equality

Inequality

Step-by-step explanation:

3 0
3 years ago
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