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fiasKO [112]
3 years ago
15

A.

Mathematics
1 answer:
vfiekz [6]3 years ago
5 0
Find, days Lydia takes to make 86 bracelets
+ number of bracelets she haven’t made: 86-14=72
+ days it take to Finnish: 72/3=24
So, she needs 24 days to make 86 bracelets
You might be interested in
. If p = -5, q = 4, then find pq - (p + q).​
bija089 [108]

Answer:

-19

Step-by-step explanation:

Given :

p = -5

q = 4

Putting the value of p and q in equation pq - (p + q)

=> (-5)(4) - [(-5)+4]

=> -20 - [ -5+4 ]

=> -20 - [-1 ]

=> -20+1

=> -19

5 0
3 years ago
Name three points in the diagram that are not collinear.
ehidna [41]

Answer:

b

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
How to solve<br> 3/5e-6=-2/5(e-10)-7
kaheart [24]
\frac{3}{5} e-6=- \frac{2}{5}(e-10)-7
\frac{3}{5} e-6=- \frac{2}{5} e- \frac{2}{5}*(-10) -7 \\  \frac{3}{5} e-6=- \frac{2}{5} e+ \frac{20}{5} -7= \frac{3}{5} e-6=- \frac{2}{5} e+4 -7= \\
\frac{3}{5}e-6=- \frac{2}{5} e-3      /*5
3e-18=-2e-15        /+15
3e-3=-2e            /+2e
5e-3=0              /+3
5e=3
e=\frac{3}{5}
6 0
3 years ago
I need help this question?
Masja [62]

Answer:

Correct answer:  F. graph F or x ∈  |-5 ; 5| (including endpoints)

Step-by-step explanation:

Let us first define the absolute value:

| x |  = 1. { x with condition x ≥ 0 }

    or 2. { - x with condition x < 0 }

This is a linear inequality

1.  x ≤ 5  ∧  x ≥ 0   ⇒   0 ≤ x ≤ 5  or interval x ∈  |0 ; 5| (including endpoints)

2. - x ≤ 5  when we multiply both sides of the equation by -1 we get:

x ≥ -5 ∧ x < 0  ⇒  -5  ≤ x < 0 or interval x ∈  |-5 ; 0) (including -5)

The solution to this linear inequality is the union of these two intervals:

x ∈  |-5 ; 0) ∪ |0 ; 5|  ⇒ x ∈ |-5 ; 5| (including endpoints)

x ∈ |-5 ; 5| (including endpoints)

God is with you!!!

7 0
4 years ago
Simplify $\frac{-10a^2b^4}{5a^{-9}b^{-5}}$
r-ruslan [8.4K]
\bf ~~~~~~~~~~~~\textit{negative exponents}&#10;\\\\&#10;a^{-n} \implies \cfrac{1}{a^n}&#10;\qquad \qquad&#10;\cfrac{1}{a^n}\implies a^{-n}&#10;\qquad \qquad &#10;a^n\implies \cfrac{1}{a^{-n}}&#10;\\\\&#10;-------------------------------\\\\&#10;\cfrac{-10a^2b^4}{5a^{-9}b^{-5}}\implies \cfrac{-10}{5}\cdot \cfrac{a^2b^4}{a^{-9}b^{-5}}\implies \cfrac{-2}{1}\cdot \cfrac{a^2a^9b^4b^5}{1}&#10;\\\\\\&#10;-2a^{2+9}b^{4+5}\implies -2a^{11}b^9
5 0
3 years ago
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