Answer:
(negative 7 x + 4)(negative 7 x minus 4)
Step-by-step explanation:
Consider two real numbers a and b. A difference of squares involving a and b is usually given as;
The difference of the two squares above can be factored to yield;
This implies that in order to have the difference of squares we must have a product involving the difference and the sum of the numbers.
The expression;
(negative 7 x + 4)(negative 7 x minus 4) can also be written as ( 4 - 7x) ( -4 - 7x)
( 4 - 7x) ( -4 - 7x) = ( 4 - 7x) (-1( 4 + 7x)) = -1 *( 4 - 7x) ( 4 + 7x)
Expanding the last expression yields;
-1 (16 + 28x -28x - 49x^2) = -1 (16 - 49x^2) = 49x^2 - 16 which is in deed a difference of squares
Total points is 23
<u>Explanation:</u>
Given:
Total number of baskets = 18
Let the number of 2 pointer shots be x
Let the number of 3 pointer shots be y
According to the question:
-1
The equation can be formed as:
On solving this equation further we get:
Putting the value of x = 4 in equation 1
Points for 2 pointer shots = 4 X 2
= 8
Points for 3 pointer shots = 5 X 3
= 15
Total points = 8 + 15
= 23
Therefore, total points is 23
It is given that:
Circle 1 has center (−4, −7) and a radius of 12 cm.
Circle 2 has center (3, 4) and a radius of 15 cm.
Two circles are said to be similar if by some translation and dilation it could be placed over the other to form the same circle.
The circles are similar because the transformation rule ( x,y ) → (x+7,y+11) can be applied to Circle 1 and then dilate it using a scale factor of 5/4
( Since, as the center of circle 1 is (-4,-7)
so,
(-4+7,-7+11) → (3,4)
( Since, the radius of circle 1 is 12 and that of circle 2 is 15 cm.
so, let the scale factor be k .
that means :
)
Answer:
Option C - Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
Step-by-step explanation:
We are given;
n = 15
t-value = 1.66
Significance level;α = 0.05
So, DF = n - 1 = 15 - 1 = 14
From the one-sample t - table attached, we can see that the p - value of 0.06 at a t-value of 1.66 and a DF of 14
Now, since the P-value is 0.06,it is greater than the significance level of 0.05. Thus we do not reject the null hypothesis. We conclude that there is not sufficient evidence that the true diameter differs from 0.5 in.
