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scoray [572]
3 years ago
13

1,003,498 word form Please my

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

one million - three thousand - four hundred - ninety eight

You might be interested in
20 pts + BRAINLIEST
exis [7]
N 4
Part A
a) 7a+2a+3b-------> (7a+2a)+3b-----> 9a+3b
b) 8p+2p-7q-------> (8p+2p)-7q------> 10p-7q
c) 9a²+2a²+5a-----> (9a²+2a²)+5a---> 11a²+5a
d) 5ab+2ab-7a-----> (5ab+2ab)-7a---> 7ab-7a
e) 7+2a-a------> 7+(2a-a)-----> 7+a
f) 2a+3b+5a+2b---> (2a+5a)+(3b+2b)-----> 7a+5b
g) 6p+2q-4p+4q---> (6p-4p)+(2q+4q)----> 2p+6q
h) 9m²+7m²+5m-3m-----> (9m²+7m²)+(5m-3m)------> 16m²+2m
i) 7a²+4a-2a²+7----> (7a²-2a²)+4a+7------> 5a²+4a+7
j) 10p-3q+p-7p-----> (10p+p-7p)-3q------> 4p-3q
k) 3x²+2x-2x²+3x+x²---> (3x²-2x²+x²)+(2x+3x)-------> 2x²+5x

N 5
a) If Julie buys 4 boxes and Michelle buys 7 boxes
in total 
(4+7)*n-----> 7*n
the answer Part 5 a) is 7*n

b) (5kg+3kg)*b---------> 8*b
the answer Part 5 b) is 8*b

c) (3+4+6)*x-------> 13*x
the answer Part 5 c) is 13*x

N 6
a) L+B+L+B------> (L+L)+(B+B)-----> 2*L+2*B-----> 2*(L+B)

b)
1) L+L+L-----> 3*L
2) 5+5+5+5-----> 5*4
3) x+x+y-----> 2*x+y
4) a+b+c+c----> a+b+2*c

N 7
a) s+s+s+s+s+s+s+s------> 8*s
b) t+t+t+t-----> 4*t
c) t+t+t+t+t+t----> 6*t
d) t+t+t+s+s----> 3*t+2*s
e) t+t+t+s+s+s----> 3*t+3*s
f) 12*t+6*s
g) 10 squares+12 triangles-----> 10*s+12*t
h) n squares+m triangles-----> n*s+m*t



7 0
4 years ago
I have one more question this is the last one im asking and again don't just put any old answer or youl be reported.
ladessa [460]

Answer:

multiply

Step-by-step explanation:

pls don't report if I a wrong!

5 0
3 years ago
Read 2 more answers
Need help answering these questions please.
Ksivusya [100]
-You would do $3.50(3)
-Which is $12.50
7 0
4 years ago
How many 5/8th are in 1?
Tom [10]
You can get only one 5/8 In 1
8 0
3 years ago
Please help! Will mark the brainliest!
sesenic [268]
The domain is the set of all possible x-values which will make the function valid.
f(x) = \frac{3}{x-2} \ \ \ \ , \ g(x) =  \sqrt{x-1}
For the given function :
The denominator of a fraction cannot be zero
The number under a square root sign must be Non negative


(a)

(1) The domain of f ⇒⇒⇒ R - {2}

Because ⇒⇒⇒  x - 2 = 0  ⇒⇒⇒ x = 2

(2) The domain of g ⇒⇒⇒ [1,∞)
Because: x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1

(3) f + g = \frac{3}{x-2} + \sqrt{x-1}
The domain of (f+g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1


(4) f - g = \frac{3}{x-2} - \sqrt{x-1}
The domain of (f-g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1


(5) f * g = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)}
The domain of (f*g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1

(6) f * f = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}
The domain of ff ⇒⇒⇒ R - {2}

Because ⇒⇒⇒  x-2 = 0  ⇒⇒⇒ x = 2

(7) \frac{f}{g} =   \frac{\frac{3}{x-2} }{ \sqrt{x-1} } =  \frac{3}{(x-2) \sqrt{x-1}}
The domain of (f/g) ⇒⇒⇒ (1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 > 0 ⇒⇒⇒ x > 1

(8) \frac{g}{f} =  \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } =  \frac{1}{3} (x-2) \sqrt{x-1}
The domain of (g/f) ⇒⇒⇒ [1,∞) - {2}
Because: x - 2 = 0 ⇒⇒⇒ x = 2   and   x - 1 ≥ 0  ⇒⇒⇒ x ≥ 1
===================================================
(b)


(9) (f + g)(x) = \frac{3}{x-2} + \sqrt{x-1}


(10) (f - g)(x) = \frac{3}{x-2} - \sqrt{x-1}


(11) (f * g)(x) = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)}


(12) (f * f)(x) = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}


(13) \frac{f}{g} =   \frac{\frac{3}{x-2} }{ \sqrt{x-1} } =  \frac{3}{(x-2) \sqrt{x-1}}


(14) (\frac{g}{f})(x) =  \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } =  \frac{1}{3} (x-2) \sqrt{x-1}



7 0
3 years ago
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