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Ierofanga [76]
1 year ago
7

The area of a rectangle is 1,357 square feet. The length is 59 feet. What is the width?

Mathematics
2 answers:
Marysya12 [62]1 year ago
8 0

Answer:

Step-by-step explanation:

23

dangina [55]1 year ago
3 0

Answer:

w = 23

Step-by-step explanation:

a = l × w

The area and length are given, so you can substitute them in.

1357 = 59 × w

Solve for w:

w = \frac{1357}{59}

w = 23

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Which situation could be represented by the graph?
Y_Kistochka [10]

Answer:

A program at a community college can be completed in no fewer than 8 months, but must be completed in less than 11 months.

Step-by-step explanation:

1. The closed dot means the number is equal to and the open dot means the number is less/ greater than

2. The line goes right from the 8 with a closed dot, making x greater than or equal to 8

3. The line goes left from the 11 with an open dot, making x less than ll

4. This means the number must be equal to or greater than 8, and less than 11

9 0
2 years ago
Read 2 more answers
If trapezoid JKLM with vertices) (3, 4), K (6,4), L (8, 1) and M (1,1) is rotated 270 degrees counterclockwise, what are the
Stels [109]

Answer:

J' = (4,-3)

Step-by-step explanation:

Given

J = (3, 4)

K =(6,4)

L = (8, 1)

M = (1,1)

Rotation = 270CCW

Required

Determine the new coordinate of J

From rules of rotation,

When a point (x,y) is rotated 270 degrees CCW;

The new point becomes (y,-x)

Considering point J

J = (3, 4)

This means

(x,y) = (3,4)

Where x = 3 and y = 4

Using the above rotation rule of

(x,y) -> (y,-x)

The coordinates of J' becomes

J' = (4,-3)

7 0
1 year ago
Joe walks on a treadmill at a constant rate. The equation below describes the relationship between t, the time he walks in hours
STALIN [3.7K]

Answer: A

Step-by-step explanation:

CORRECTED QUESTION

Joe walks on a treadmill at a constant rate. The equation d=4t describes the relationship between t, the time he walks in hours, and d, the distance he walks in miles. Which of the graph best describes the relationship?

The equation d=4t is a linear equation.

When t=1, d=4 X 1 =4

When t=2, d=4 X 2 =8

The correct graph that satisfies this pair is A (as seen in the diagram)

4 0
2 years ago
For which discriminant is the graph possible<br> b2-4ac=0<br><br> b2-4ac=-1<br><br> b2-4ac=4
VARVARA [1.3K]

Answer:

The graph is possible for b^2-4ac=4

Step-by-step explanation:

we know that

The discriminant of a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

D=b^2-4ac

If D=0 the quadratic equation has only one real solution

If D>0 the quadratic equation has two real solutions

If D<0 the quadratic equation has no real solution (complex solutions)

In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)

so

b^2-4ac > 0

therefore

The graph is possible for b^2-4ac=4

4 0
1 year ago
Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.
xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form p(t)=at^2, where a is in ℝ.

2.Let W2 be the set of all polynomials of the form p(t)=t^2+a, where a is in ℝ.

3. Let W3 be the set of all polynomials of the form p(t)=at^2+at, where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

p_1 (t) = at^2, p_2(t)=bt^2.

We must check that p_1+p_2(t) is in W1.

Note that

p_1(t)+p_2(t) = at^2+bt^2  = (a+b)t^2

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say p_1(t) = a(t^2+t), p_2(t)= b(t^2+t). Then

p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 p(t) = at^2and a real scalar b. Then

bp(t) = b(at^2) = (ba)t^2).

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 p(t) = a(t^2+t)and a real scalar b. Then

bp(t) = b(a(t^2+t)) = (ba)(t^2+t).

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.  

6 0
1 year ago
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