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dsp73
3 years ago
7

Mrs. Hinely grows roses. There are 6 roses on each of her 10 rose bushes. How many roses in all are on Mrs. Hinely's rose bushes

?
Mathematics
2 answers:
diamong [38]3 years ago
8 0
In order to find the total number of roses, we can use multiplication. Since there are 6 roses on each bush, we can say that the answer is: 6+6+6+6+6+6+6+6+6+6. Since multiplication is a way to quickly add the same number, we can use 6*10. This becomes 60. 

Good luck :D

9966 [12]3 years ago
7 0
60 roses on mrs hinelys rose bushes!
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Mercury is a naturally occurring metal that can be harmful to humans. The current recommendation is for humans to take in no mor
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Answer:

A.) A person weights 80 kg can consume Haddock fish = 149 grams

B.) A person weights 80 kg can consume Swordfish fish = 8 grams

C.) A person weights 80 kg can consume Tuna fish = 23 grams

D.) A person weights 80 kg can consume Snapper fish = 50 grams

Step-by-step explanation:

Given - Mercury is a naturally occurring metal that can be harmful to humans. The current recommendation is for humans to take in no more than 0.1 microgram for every kilogram of their weight per day. Fish generally carry high levels of mercury, although certain fish have higher mercury content than others. Fish, however, are healthy sources of many other nutrients, so nutritionists recommend keeping them in the human diet. The figure below shows the average mercury content of several types of fish.

To find - If a person weighs 82 kg, how much of each fish can they safely consume? round to the nearest gram.

A.Haddock

B.Swordfish

C.Tuna

D.Snapper​

Proof -

Firstly, determine how many micro grams of mercury a person can consume who weighs 82 kilograms.

⇒82 × 0.1 =  8.2 micro grams.

Now,

a.)

For Haddock fish-

It consumes 0.055 micro grams of mercury

So,

A person can consume fish = 8.2 ÷ 0.055

                                              = 149.09 ≈ 149 grams

So,

A person weights 80 kg can consume Haddock fish = 149 grams

b.)

For Swordfish fish-

It consumes 0.995 micro grams of mercury

So,

A person can consume fish = 8.2 ÷ 0.995

                                              = 8.24 ≈ 8 grams

So,

A person weights 80 kg can consume Swordfish fish = 8 grams

c.)

For Tuna fish-

It consumes 0.350 micro grams of mercury

So,

A person can consume fish = 8.2 ÷ 0.350

                                              = 23.43 ≈ 23 grams

So,

A person weights 80 kg can consume Tuna fish = 23 grams

d.)

For Snapper fish-

It consumes 0.165 micro grams of mercury

So,

A person can consume fish = 8.2 ÷ 0.165

                                              = 49.697 ≈ 50 grams

So,

A person weights 80 kg can consume Snapper fish = 50 grams

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3 years ago
Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!
blagie [28]

<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

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3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

5) \overline{BE} \cong \overline{BE} (reflexive property)

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

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1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

2) \overline{AC} \cong \overline{AC} (reflexive property)

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

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6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

5) \overline{AD} \cong \overline{AD} (reflexive property)

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

7) \angle BDA \cong \angle CDA (CPCTC)

8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

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The answer is 54,550 more.
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