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Nadya [2.5K]
3 years ago
13

A triangle has height 5 cm and base length 8 cm find it’s area

Mathematics
1 answer:
vodka [1.7K]3 years ago
4 0

Answer:

A = 20 cm^2

Step-by-step explanation:

We can find the area of a triangle from

A = 1/2 bh

A = 1/2 8*5

A = 20 cm^2

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Given matrices S and T below, which statement is true?
harina [27]
The matrices are
S =(4 11                           T= ( -8 11
      -3  -8)                                 3  4 )
Inverse of a matrix is a matrix derived from another matrix such that if you pre- multiply it with the original matrix you get a unit matrix.
if we multiply S and T
 ST will be 
        ( 4 11    ×      (-8 11   = ( 1 0
          -3 -8)           -3 -4)         0 1)
and also TS
          ( -8 11      ×     (4 11  = ( 1 0
             -3 -4)            -3 -8)       0 1)
therefore, matrices S and T are inverses of each other because ST = TS= I 



.
8 0
3 years ago
Read 2 more answers
NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions
Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector \vec{v} in some vector space \mathbb R^n we have

\vec{v} = \langle a,b\rangle

This is the component form. I don't like that way. It is probably used in high school, but

\vec{v} =  \begin{pmatrix} a\\ b\\ \end{pmatrix}

is preferable because the inner product on \mathbb R^n is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as \vec{v} = 2i+2j

4.

For this question, I think you meant

vectors

\vec{u_1} = (-8, 12)

\vec{u_2}  = (13, 15)

Once

\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}

Considering that the dot product is

\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76

and the norm of \vec{u_1} is ||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}

and the norm of \vec{u_2} is ||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}

Thus,

\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}

\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)

3 0
3 years ago
Solve for z.<br><br> z12≤3<br><br> z≥36<br><br> z≤36<br><br> z≥4<br><br> z≤4
erastova [34]

Answer:

Your answer is correct... it's no.3 ... but if you multiply a inequality by a negative value the inequality sign changes.. like if you habe greater than it'll be less than... Keep that in mind too

5 0
3 years ago
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BC =<br> Round your answer to the nearest hundredth.<br> ?<br> C<br> B<br> 50°<br> 7<br> А
umka21 [38]

Answer:

The answer is 8.34

U have to use the SOH CAH TOA method

In this case I used tan

so,

tan 50= BC/7

solve for BC= 7* tan 50= 8.34

6 0
3 years ago
Hurry and answer please
arlik [135]

It's c (48)

because if you divide all of them its 3

5 0
3 years ago
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