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3 years ago

What is the area of the triangle? please helpppp

Mathematics

2 answers:

vladimir2022 [97]3 years ago

6 0

where did the you hide the triangle?

Ilya [14]3 years ago

5 0

Where is the picture of the triangle?

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jill bikes for 1/5 of an hour at 16 mph and then walks for 3/7 hour at 5 miles per hour how much farther did she bike then walk

Montano1993 [528]

Ans: 1.1 miles more

Speed = distance/time

Time is 1/5 of an hour

1/5 of 1hr = 1/5 or 0.2 hrs

Speed = 16 mph

So, we're after distance.

Distance = speed x time

= 16mph x 0.2

= 3.2 miles

Walks for 3/7 hour at 5 miles per hour :

Time & speed given, distance is what we're after.

(3/7 of 1hr is 3/7)

Distance = speed x time

= 5mph x 3/7

= 15/7 miles or 2.14285...miles

So, distance is about 2.1 miles to 1 d.p

Thus, 3.2 - 2.142... (the full ans) = 1.057... miles

Well, he biked 1.06 miles or 1.1 miles (depending on how you want to round it) more than walking.

Hope this helps!

7 0

2 years ago

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Suppose an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline that

lapo4ka [179]

Answer:

this is very confusing to understand that you want.

Step-by-step explanation:

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3 years ago

PLSSSS HELPPPPPPPPPPPPPPP

g100num [7]

Answer:

i can hardly see that

Step-by-step explanation:

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3 years ago

5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

3 0

3 years ago

Complete the statement. If a transversal intersects two parallel lines, then ___.

almond37 [142]

Answer:

alternate interior angles are always congruent.

Step-by-step explanation:

- Consecutive [Same-side] exterior angles are always supplementary.

- Consecutive [Same-side] interior angles are always supplementary.

- Corresponding angles are always congruent.

I am joyous to assist you anytime.

4 0

3 years ago

Read 2 more answers