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VLD [36.1K]
3 years ago
11

Can someone please help me with this​

Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Answer:

y = 5

z = 55

Step-by-step explanation:

90 - 35 = 55

z = 55

7y = 35

\frac{7y}{7} = \frac{35}{7}

y= 5

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2x^2 + 3x - 12 when x = 5 help pls
cluponka [151]

\huge\textsf{Hey there!}

\mathsf{2x^2 + 3x - 12}

\mathsf{= 2(5)^2 + 3(5) - 12}

\mathsf{5^2}

\mathsf{= 5 \times 5}

\mathsf{\bf = 25}

\mathsf{2(25) + 3(5) - 12}

\mathsf{2(25)}

\mathsf{= \bf 50}

\mathsf{3(5)}

\mathsf{= \bf 15}

\mathsf{= 50 + 15 -  12}

\mathsf{50 + 15}

\mathsf{= \bf 65}

65 - 12

\mathsf{= \bf 53}

\boxed{\boxed{\huge\textsf{Answer: \bf 53}}}\huge\checkmark

\large\textsf{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

5 0
3 years ago
Read 2 more answers
5,977,243 x 457,766 pls answer
9966 [12]

Answer:

a number

Step-by-step explanation:

5 0
3 years ago
15, -19, 23, -27.. how would you write that in sigma notation?
sertanlavr [38]

Answer: i think its zero

Step-by-step explanation:

or u can just look it up on an sigma math calculator and that should give u the answer

3 0
2 years ago
PLEASE HELP!!!!! How many 4-digit numbers divisible by 5, all of the digits of which are even, are there?
saul85 [17]

Answer:

I assume you know Arithmetic Progression .

so, we have to find the first and last 4-digit number divisible by 5

first = 1000 ,  last =  9990

we have a formula,   a_{n} = a + (n-1)d

here, a_{n} is the last 4-digit number divisible by 5.

n is the number of 4-digit even numbers divisible by 5

d is the common difference between the numbers, which is 10 in this case

a is the first 4-digit number divisible by 5

9990 = 1000 + (n-1)*10

899 = n-1

n = 900

Hence, there are 900 4-digit even numbers divisible by 5

7 0
3 years ago
Helpppp will mark brainliest.
Tju [1.3M]
8 ^ (-2) = 1 / 64
<span>4 ^ (-3) = 1 / 64</span>
4 0
3 years ago
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