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Note that   x^2 -7x + 10 =   (x - 2) ( x - 5)

Multiply through by the factorized form and we  have

5x + 16 =   A ( x - 5)  + B ( x - 2)

5x +  16 =   (A + B)x  +  ( -5A - 2B)

Equating terms we  see that

A + B =   5

The base of the  triangle =  5

To find the height

Area = (1/2) (5)  (height)

15  = (5/2)  (height)

15 (2/5)  = height

6  = height........this is the y interecept

Intercepts are  (-5,0) and (0 , 6)

Slope of line =  ( 6 - 0)  / ( 0 - -5)  =  6 / 5

This simplifies to \(1 + \sqrt[3]{3}\)

        x    y

1 = (36, 1)
2 = (9, 8)
3 = (4, 27)

x ==36 + 9 + 4 ==49

2 * 0.5^6 - 0.5^7 =  0.0234375

There seems to be a notational or transcription error in the question.

The statement that the first row of the matrix A is the vector u implies that u is a row vector.

(Similarly, the implication is that v is a row vector.)

We are not given the dimension of u and v, but assuming that it is two, A will be a 2 by 2 matrix.

In that case, the products Au and Av do not exist, for the products to exist u and v would need to be column vectors.

The question would make sense if we were given that

\(\displaystyle \text{row}_{1}(A)=u^{\text{T}},\)

(and similarly for v),

or, calculate

\(\displaystyle \text{A}u^{\text{T}},\)

(If u is a row vector.)

Question: Solve the inequality \(4t^2 \le 9t-2\)

We rearrange it as follows:
\(4t^2-9t+2 \le 0\)

Then, factor this quadratic equation:
\((4t-1)(t-2) \le 0\)

The critical values (also called endpoints) are: \(t_1=\frac{1}{4},t_2=2\)

So, if we sketched the quadratic \(4t^2-9t+2\), it has x-intercepts (1/4 and 2).

And we want the region below the x-axis (As it is less than 0).

Hence, the region between 1/4 and 2.

Therefore: \(\frac{1}{4} \le t \le 2\)

Here is a sketch: (Using desmos.com)

I hope this helps.

Here's my best attempt at this one ....

There are   C(9,4)  = 126 possible sets

If we consider the sets  whose  product is NOT divisible by 4, we have the folowing  cases

Case 1  -   the set contains  four of the following six integers .... { 1, 2 , 3, 5, 7 , 9 }

Total sets possible =  C(6,4)  =  15

Case 2  - the set contains four of the following six integers  ....{1, 3 , 5, 6, 7  ,9 }

Total sets possible = C(6,4)  =  15

Total sets whose product is divisible by 4   =  126 - 2(15)  =  96

P ( that the set product is divisible by 4 )  =  96 / 126  =  16 / 21 

[ I'd like someone to  check  this answer.....!!!!  ]

what value of k

Hello Guest!

\( \)

\(-1+kx = 3y\\ k=\frac{3y+1}{x}=\frac{3\cdot (-8)+1}{4}\\ \color{blue}k=-5.75\)

  !

0

It cannot land back at point A, not after 3 moves

The domain will be all real numbers except for 2 values. Can you find them? (Hint; They will be a and b!)

The range will also be all real numbers as well except for one value. Try inverting the x and y values and isolating the new y value! (Hint: that number would be c!)

The dot product of u and v is equal to ||u||* ||v||*cos 60 =  3. (Prove this before you start using it!)

row_1(Au) = \(\mathbf{u \cdot u}\) = 9

row_2(Au) = Dot product of u and v, equaling 3

Au = <9, 3>

Similarly:

row_1(Av) = 3

row_2(Av) = v \cdot v = 4

Av = <3, 4>

They answered it when this post was asked

oh, builderbot, it's so tempting to call Q a right angle.  It looks like

it's a right angle, and that would make side PR come out even. 

Let k be the largest integer such that 343/7^k is an integer. What is the remainder when 343/7^k  is divided by 7?  

The value of k = 3.  This is because 7³ is 343, therefore 343/7³ = 1. 

If the exponent is any value larger than 3, then the quotient is less than 1. 

I don't understand what the second part of the question is asking for.  Sorry. 

_.

Something like this (with the placement of D  inexact )

A                                D

8                                         5sqrt (2)

B               20                                   C

Draw  AC

AC  = sqrt [ 20^2 + 8^2 ]  =  sqrt [464 ]   =  4sqrt (29)

tan (ACB)  = 8/20  = 2/5

arctan (2/5)  = measure of angle ACB

Measure  of  angle  ACD = measure of angle BCD  - measure of angle ACB =    45  - arctan(2/5)

Area of figure = area of right triangle ABC  +  area of triangle ACD  = 

(1/2) (AB) (BC)  + (1/2) (AC) ( DC) sin ( ACD)  =

(1/2) (8) (20)  + (1/2) (4sqrt (29) ) ( 5sqrt (2) ) sin  [ 45  - arctan (2/5) ]   =

80  +  10 sqrt ( 58) * sin [45 - arctan (2/5) ]  =     110  units^2 

I dont know what 'total turns' means but if the first and second ones loses on their first and second go then the game is over. so that is 5 goes in total.