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Aloiza [94]
3 years ago
5

What is the perimeter of the rectangle? A 30 cm B. 60 cm C. 200 cm D. 400 cm

Mathematics
2 answers:
Olin [163]3 years ago
5 0

Answer:

60 cm

B.)60 cm

Ratling [72]3 years ago
4 0

Answer:

B: 60 cm

Step-by-step explanation:

Because the perimeter you have to add all sides

so it would be 20+10+20+10 = 60

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Suri's age is 4 less than 3 times her cousin's age. Suri is 17 years old. Which method can be used to find c, her cousin's age?
Mila [183]

Answer:

7 years

Step-by-step explanation:

Suri's age=17 years

Her cousin's age =c

Suri's age is 4 less than 3 times her cousin's age

4 less than 3 times her cousin's age means subtract 4 from 3 times her cousin's age

17=3c - 4

Find c which is her cousin's age

Add 4 to both sides

17+4=3c - 4 + 4

17+4=3c

21=3c

Divide both sides by 3

21/3=3c/3

7=c

Therefore,

C= 7

Her cousin's age = 7 years

4 0
3 years ago
The equation of a line is y = -4x + 10<br> What is the y-intercept of the line
Vikentia [17]

Answer:

10

Step-by-step explanation:

the formula y = mx +c

the y is obviously, y.

m, also means the gradient has a value of -4

c is the y-intercept, so the value of c is 10.

4 0
2 years ago
Which expressions are equivalent to -6 + 3(2+ (-4t)?
lilavasa [31]
I can assure you that it is A
8 0
3 years ago
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Whats 8+0.3 in standard form and word form
Rudik [331]
8 + 0.3  = 

Standard . . .  8.3

Words . . . "eight and three tenths".
3 0
3 years ago
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Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
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