The answer is d x-7 because it's seven less than a number
<h3>I'll teach you how to solve sqrt 150</h3>
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sqrt 150
Prime factorization:
sqrt 2*3*5^2
Apply radical rule:
sqrt 2*3 sqrt 5^2
Apply radical rule:
sqrt 2*3 *5
Multiply the numbers:

Your Answer Is (C) 
plz mark me as brainliest :)
Answer:
-6
Step-by-step explanation:
Given that :
we are to evaluate the Riemann sum for
from 2 ≤ x ≤ 14
where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.
The Riemann sum can be computed as follows:

where:

a = 2
b =14
n = 6
∴



Hence;

Here, we are using left end-points, then:

Replacing it into Riemann equation;






Estimating the integrals, we have :

= 6n - n(n+1)
replacing thevalue of n = 6 (i.e the sub interval number), we have:
= 6(6) - 6(6+1)
= 36 - 36 -6
= -6
400% of a is the 1% of 400a
mode = 43
the mode is the value which occurs most
43 occurs 3 times, 38 twice and the others only once
Hence 43 is the mode