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Aleks [24]
3 years ago
12

A map has a scale of 4 inches :25 miles .How far are two cities on a map if they are 120 miles apart?

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

19.2 in

Step-by-step explanation:

Set up a proportion to solve this...

4 inches is to 25 miles as x inches is to 120 miles

4/25 = x/120         now cross multiply to and solve for x

480 = 25x

 480/25 = x

     19.2 = x

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6. Juan recibe de su papá, quien tiene un negocio de maquinitas, su mesada semanal de $62.000 en monedas de $200,
Y_Kistochka [10]

Answer:

15100

Step-by-step explanation:

7 0
3 years ago
Which of the following is an equation for the plane through the point (1, 1, −1) and parallel
Bas_tet [7]

The given plane, x-y+z=3, has normal vector \vec n = \langle1,-1,1\rangle. Any plane parallel to this one has the same normal vector.

Let (x,y,z) be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is

\langle x,y,z\rangle - \langle 1,1,-1\rangle = \langle x-1, y-1, z+1 \rangle

and this is perpendicular to \vec n.

So the equation of the plane is

\langle x-1, y-1, z+1 \rangle \cdot \vec n = 0 \implies (x-1) - (y-1) + (z+1) = 0 \implies \boxed{x - y + z = -1}

or equivalently,

\boxed{-x + y - z = 1}

6 0
2 years ago
Sarah Meeham blends coffee for Tasti-Delight. She needs to prepare 160 pounds of blended coffee beans selling for $4.97 per poun
jok3333 [9.3K]
Given:
let h be the high quality bean
let c be the cheaper bean

h + c = 160
6h + 3.25c = 160*4.97
6h + 3.25c = 795.20

h = 160 - c
6(160 - c) + 3.25c = 795.20
960 - 6c + 3.25c = 795.20
-2.75c = 795.20 - 960 
-2.75c = -164.80
c = -164.80 / -2.75
c = 59.92 or 60 lbs

h = 160 - c
h = 160 - 60
h = 100 lbs

Sarah should blend 60 lbs of cheap coffee bean and 100 lbs of high quality coffee bean.
3 0
3 years ago
Find the value of f(g(-2))
PSYCHO15rus [73]

Answer:

The correct ans is 3.

g(-2) = 2(-2) + 5

=-4 + 5 = 1

f(g(-2)) = 4 - (1)^2

= 4 - 1

=3

Step-by-step explanation:

4 0
3 years ago
Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that A=J^{-1}AJ. Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
3 years ago
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