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Neko [114]
3 years ago
8

The product of a number and 7

Mathematics
2 answers:
musickatia [10]3 years ago
6 0
Product: Multiplication. therefore, 7n 
Zinaida [17]3 years ago
4 0
What is the number? Or do you want us to give you the number?  If that is the case, choose 3 * 7, or maybe 6 * 7.
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Answer is 239,800.

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Solve the following equation then place the correct number in the box provided. X/3 - 4 = 10
ivann1987 [24]
X = 42

x/3 - 4 = 10
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Line I is parallel to line m. If the measure of <6 is 75, what is the measure of <2?
Tamiku [17]

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the answer is 25 in this question hope it helped

5 0
2 years ago
Read 2 more answers
What is the length of AB?
Xelga [282]

Answer:

<em>AB = 7.35 cm</em>

Step-by-step explanation:

From the attachment,

In ΔDEF,

DF = GH-(GD+FH) = 6 - (2+3) = 1 cm

DE = 2+3 = 5 cm (sum of two radius)

Applying Pythagoras theorem,

EF=\sqrt{DE^2-DF^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt6

In ΔCDI,

DI = GH-(GD+IH) = 6 - (2+1.5) = 2.5 cm

CD = 2+1.5 = 3.5 cm (sum of two radius)

Applying Pythagoras theorem,

CI=\sqrt{CD^2-DI^2}=\sqrt{3.5^2-2.5^2}=\sqrt6

AB = EF + CI = 2\sqrt6+\sqrt6=3\sqrt6=7.35\ cm

6 0
3 years ago
Read 2 more answers
1.Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by
Genrish500 [490]

If x is a real number such that x3 + 4x = 0 then x is 0”.Let q: x is a real number such that x3 + 4x = 0 r: x is 0.i To show that statement p is true we assume that q is true and then show that r is true.Therefore let statement q be true.∴ x2 + 4x = 0 x x2 + 4 = 0⇒ x = 0 or x2+ 4 = 0However since x is real it is 0.Thus statement r is true.Therefore the given statement is true.ii To show statement p to be true by contradiction we assume that p is not true.Let x be a real number such that x3 + 4x = 0 and let x is not 0.Therefore x3 + 4x = 0 x x2+ 4 = 0 x = 0 or x2 + 4 = 0 x = 0 orx2 = – 4However x is real. Therefore x = 0 which is a contradiction since we have assumed that x is not 0.Thus the given statement p is true.iii To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false.Here r is false implies that it is required to consider the negation of statement r.This obtains the following statement.∼r: x is not 0.It can be seen that x2 + 4 will always be positive.x ≠ 0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with x2 + 4.∴ x x2 + 4 ≠ 0⇒ x3 + 4x ≠ 0This shows that statement q is not true.Thus it has been proved that∼r ⇒∼qTherefore the given statement p is true.

8 0
2 years ago
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