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3 years ago

Plzzzzzz helpppp asap why the hell do peaple not answer

Mathematics

1 answer:

o-na [289]3 years ago

6 0

Answer:

The base is 10 squares by 5 squares

Step-by-step explanation:

Well 18m is three squares so 18/3 = 6m which is 6m per square

The manor is 60m by 30m

so 60/6 = 10

and 30/6= 5

The base is 10 squares by 5 squares

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If the triangles are similar, only a, only b, both, or neither

Zarrin [17]

Answer:

Step-by-step explanation:

0.9/1=0.9

13.5/15=0.9

8/5=1.6

3 0

3 years ago

He sum of 1/6, 2/3, and 1/4 is

Alborosie

The LCD here is 12. Th,us, 1/6, 2/3 and 1/4 become:

2/12, 8/12 and 3/12.

Their sum is 13/12, or 1 1/12.

7 0

3 years ago

Read 2 more answers

Choose the equation below that represent the line passing through the point (-2,-3) With a slope of -6

Gnom [1K]

Y+3= -6 (x+2) is the correct equation

5 0

4 years ago

Read 2 more answers

6. Amy bought some used books for $4.95. She paid $0.50 each for some books and $0.35 each for the others. She bought fewer than

Sauron [17]

Amy bought 5 books for $0.50 and 7 for $.30, a total of 12
WORK
5 x .50 = 2.5
4.95 - 2.5 = 2.45
2.45 divided by .35 = 7

3 0

3 years ago

In Exercises 1–4, let S be the collection of vectors cxyd in R2that satisfy the given property. In each case, either prove thatS

Artyom0805 [142]

Answer:

S is not the subspace of $R^2$

Step-by-step explanation:

Let us suppose two vectors u and v belong to the S such that the property of xy≥0 is verified than

$u=\left[\begin{array}{c}-1 \\0\end{array}\right]$

$v=\left[\begin{array}{c}0 \\1\end{array}\right]$

Both the vectors satisfy the given condition as follows and belong to the S

$x_u y_u=-1\times 0=0 \geq 0\\x_v y_v=1\times 0=0 \geq 0\\$

Now S will be termed as subspace of R2 if

u+v also satisfy the condition
ku also satisfy the condition

Taking u+v

$u+v=\left[\begin{array}{c}-1 \\0\end{array}\right]+\left[\begin{array}{c}0 \\1\end{array}\right]\\u+v=\left[\begin{array}{c}-1+0 \\0+1\end{array}\right]\\u+v=\left[\begin{array}{c}-1 \\1\end{array}\right]$

Now the condition is tested as

$\\x_{u+v} y_{u+v}=-1\times 1=-1$

This indicates that the condition is not satisfied so S is not the subspace of $R^2$

3 0

3 years ago