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BigorU [14]
3 years ago
5

Nancy walked 50 minutes each dayfor 4 days last week. Gillian walked 35 minutes each day for 6 days last week. How does the tota

l number of minutes that Gillian walked compare to the total number of minutes that Nancy walked?
Mathematics
2 answers:
GrogVix [38]3 years ago
8 0
Nancy:50*4=200min

Gillian:35*6=210min

Nancy walked less minutes than Gillian
KIM [24]3 years ago
7 0
Gilliam walked for 10 more minutes than Nancy
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Nady [450]

Answer:

the answer is B and C

Step-by-step explanation:

9x4=36

6x6=36

6 0
2 years ago
Answers need help!!!!!!
Gala2k [10]
<h3>Step-by-step explanation:</h3>

<u>Pythagorus theorum:</u>

9² = 3² + YZ²

81 - 9 = YZ²

√72 = YZ

∴YZ= 8.48 ≈ 8.5

<u>Trigonometry:</u>

We have adjacent as 3, and hypotenuse as 9

cos X = 3/9

X= cos∧-1 (3/9) → <em>[cos inversed]</em>

∴X= 70.5°

We have opposite as 3 and hypotenuse as 9

sin Z= 3/9

Z= sin∧-1 (3/9)→<em> [sin inversed]</em>

∴Z= 19.47 ≈ 19.5

4 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD=√2 cm.
Zielflug [23.3K]

Given that triangle ABC is right angle triangle. CD is altitude such that AD=BC

ABC is a right angle triangle so apply Pythogorean theorem

AC^{2} + BC^{2} = AB^{2}

AC^{2} + AD^{2} = AB^{2}      (Given that AD = BC)

AC^{2} + AD^{2} = 3^{2}      (Given that AB=3)

AC^{2} + AD^{2} = 9 ...(i)

ADC is a right angle triangle so apply Pythogorean theorem

AD^{2} + CD^{2} = AC^{2}

AD^{2}+(\sqrt{2})^2= AC^{2}

AD^{2}+2= AC^{2}

AD^{2}=AC^{2} -2 ...(ii)


Plug value (ii) into (i)

AC^{2} + AC^{2}-2 = 9

2AC^{2} -2=9

2AC^{2} =11

AC^{2} =\frac{11}{2}

AC=\sqrt{\frac{11}{2} }


Hence final answer is AC=\sqrt{\frac{11}{2} }

8 0
3 years ago
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The sum ∫2−2() ∫52()−∫−1−2() can be written as a single integral in the form ∫() determine and
kow [346]

We have

\displaystyle \int_{-2}^2 f(x) \, dx - \int_{-2}^{-1} f(x) \, dx = \int_{-1}^2 f(x) \, dx

so that

\displaystyle \int_{-2}^2 f(x) \, dx + \int_2^5 f(x) \, dx - \int_{-2}^{-1} f(x) \, dx \\\\ ~~~~~~~~~~~~ =  \int_{-1}^2 f(x) \, dx + \int_2^5 f(x) \, dx \\\\ ~~~~~~~~~~~~ = \boxed{\int_{-1}^5 f(x) \, dx}

6 0
1 year ago
ABO please help brailest I promise Help me please I beg you !
mash [69]
I'm just going to give the answers snce so many questions

1. 2
2. 2/(m^2)
3. 1/(16x^8)
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last graph one: third one (one that starts at 2 up from the origin)
7 0
3 years ago
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