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Lisa [10]
2 years ago
12

alt="\sqrt(1-2)+\sqrt(1+2)\\" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Kipish [7]2 years ago
8 0
1.4 or square root of 2
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Find an equation for the line below
tia_tia [17]
Y=0.5x-0.5
( u find the slope using rise/run then after finding the slope u use the equation and a point on the graph to find the y-intercept
5 0
3 years ago
Can (5z+3)(-5z-3) result in a difference of squares
castortr0y [4]

Answer:

no, it cannot

Step-by-step explanation:

a difference of square is: a² - b² = (a - b)(a + b)

looking at the expression (5z+3)(-5z-3), we see that it does not fit the criteria of the breakdown of a perfect square, as (-5z-3) has a negative <em>a</em> term (-5z)

if we FOILed (5z+3)(-5z-3) out, we would get:

-25z² - 30z - 9, which is not a difference of squares

8 0
3 years ago
By what percentage must the diameter of a circle be increased to increase its area by 50%?
katovenus [111]
The circle increase in area will be by a factor of 1.5
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5 0
3 years ago
Read 2 more answers
Please help me! I've tried so hard and just cant
torisob [31]
So... 4 folks, working for 8hours per day, do it in 15days

hmm alrite...so 4 folks working for 8hrs in 1 day, each one does 8, 4 of them, 4*8 = 32, so, they do 32 workhours every day, now for 15 days, that'd be 32*15, or 480

so, the whole job really takes 480 hours

now hmm let's see, the firm decides to increase the folks, but decrease the hours, so, 5 folks each working 6hours every day

how many workhours total? well 5*6 = 30
each is working 6hrs, 5 of them, so 30 workhours

how long will it take them if they only do 30 workhours everyday?

well the whole job is 480hours

\bf \cfrac{30hrs}{day}=480hrs\implies \cfrac{30hrs}{480hrs}=day\implies \cfrac{1}{16}=day&#10;\\\\\\&#10;\textit{so, if they do }\frac{1}{16}\textit{ daily, to complete, they'd need }\frac{16}{16}\\&#10;\\\\\&#10;\textit{for the whole, thus }16 days
5 0
3 years ago
I-Ready
Sindrei [870]

Answer:

2

Step-by-step explanation:

(f+g)(x)

f(x)= x-3, g(x)= 2x+8

3 0
3 years ago
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