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2 years ago

The product of 5 and 3 subtracted from 25

Mathematics

2 answers:

tiny-mole [99]2 years ago

8 0

The answer is 10

explanation
5x3=15
25-15=10

Gelneren [198K]2 years ago

6 0

Answer:

Step-by-step

5*3=15

25-15=10

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In the diagram below. C is an equal distance to A and B.

Fynjy0 [20]

Answer:

C(50;-20)

Step-by-step explanation:

The question is asking for the midpoint of the line.

The triangle is formed by placing a line directly from the y-axis to the x-axis.

The midpoint theorem is[ (x1+x2)/2 ; (y1+y2)/2 ].

your coordinates at A is (0;-40), because the x value is 0 on the line as shown in the diagram above point A.

Your coordinates for B are (100;0)

C[(x1+x2)/2 ; (y1+y2)/2 ]

C [(0+100)/2 ; (-40+0)/2 ]

C[ 100/2 ; -40/2]

C (50 ; -20)

7 0

2 years ago

A student walks from her house to her school. The graph shows her distance, in miles, from the school over the time in minutes,

valentina_108 [34]

ANSWER:

distance when starting her walk/ when time is 0

Step-by-step explanation:

y axis will be distance and x intercept is the point where y is 0

so represents when distance is 0.

6 0

2 years ago

(x-3)(2x+1) Can someone help me on this please

a_sh-v [17]

2x^2-5x-3

This is the answer I believe, I apologize if it’s incorrect

6 0

2 years ago

If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$