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Tanzania [10]
3 years ago
10

Recently one morning the temperature at 7:00 AM was 44 degrees Fahrenheit. If the temperature rose an average of 2 degrees per h

our and peaked at 3:00 PM, what was the temperature at 3:00 PM?
Mathematics
2 answers:
hammer [34]3 years ago
7 0
It would be 60 degrees at 3:00 PM
Arlecino [84]3 years ago
7 0

Answer: The temperature would be 60 degrees farhrenheit at 3:00 pm

Step-by-step explanation:

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I NEED HELP!!!! WITH NUMBER 9, & 10??
s344n2d4d5 [400]

Answer:

awnser one is 7 and question 2 is -10

Step-by-step explanation:

−12−2−−55=?  question one

In the first fraction, the negative numerator and negative denominator cancel each other.

Since the the second fraction is negative and you are subtracting, remove the negative sign and switch the operation to addition.

The equivalent equation is

122+55=?

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(12/2, 5/5) = 10

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

(122×55)+(55×22)=?

Complete the multiplication and the equation becomes

6010+1010=?

The two fractions now have like denominators so you can subtract the numerators.

Then:

60+1010=7010

Since 70 divided by 10 equals 7 then,

7010=7

Therefore:

−12−2−−55=7    

Question 2

−81+6−3=?

In the second fraction, move the negative sign from the denominator to the numerator and the value remains the same.

The equivalent equation is

−81+−63=?

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(-8/1, -6/3) = 3

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

(−81×33)+(−63×11)=?

Complete the multiplication and the equation becomes

−243+−63=?

The two fractions now have like denominators so you can add the numerators.

Then:

−24+−63=−303

Since -30 divided by 3 equals -10 then,

−303=−10

Therefore:

−81+6−3=−10

Solution by Formulas

Apply the fractions formula for addition, to

−81+6−3

and solve

(−8×−3)+(6×1)1×−3

=24+6−3

=30−3

=−10

7 0
3 years ago
Percy works two part-time jobs to help pay for college classes. On Monday, he works 3 hours at the library and 2 hours at the co
MrRa [10]
Percy earns a greater hourly wage of $7.50 at the library. 
6 0
4 years ago
The group is planning to build a fence around the garden. How many yards of fencing materials do they need for the fence? Show y
garri49 [273]

Answer:

They need 130 yds to build up the fence.

Step-by-step explanation:

Attached you can find a picture that will guide the explanation. We will assume that each line is supposed to be a fence. We will assume also that the diagonals that cross any subfield that has the same vegetable is unnecessary.  That is that the diagonal of the tomatoes and the watermelon are not going to be built.

Recall that the red peppers area is a right triangle. So, using pythagora's theorem, we have that

b^2 +12^2 = 15^2 so b=15^2-12^2

which gives us that b = 9.

On the same fashion, the diagonal of the spinach-celery sector is \sqrt[]{8^2+6^2}=10. Recall that the field is a rectangle, so the perimeter of the outter fence is 2*(12+6)+2*(8+9) = 70 yds. Then, the total perimeter is given by

outter fence (70)+

fence spinach-watermelon (8) +

fence red peppers-watermelon(12)+

fence red peppers-carrots (15)+

fence carrots-tomatoes (9)+

fence celery-tomatoes(6) +

fence spinach-celery(10)

= 130 yds

(if the diagonals of tomatoes and watermelon are added, the same procedure applies. watermelon diagonal is 14.42 yds and tomatoes diagonal is 10.81 yds)

6 0
4 years ago
Verify the identity. <br><br> tan x plus pi divided by two = -cot x
kogti [31]
Recall that sin(π/2) = 1, and cos(π/2) = 0,

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\bf tan\left(x+\frac{\pi }{2}  \right)=-cot(x)\\\\&#10;-------------------------------\\\\&#10;tan\left(x+\frac{\pi }{2}  \right)\implies \cfrac{sin\left(x+\frac{\pi }{2}  \right)}{cos\left(x+\frac{\pi }{2}  \right)}&#10;\\\\\\&#10;\cfrac{sin(x)cos\left(\frac{\pi }{2}  \right)+cos(x)sin\left(\frac{\pi }{2}  \right)}{cos(x)cos\left(\frac{\pi }{2}  \right)-sin(x)sin\left(\frac{\pi }{2}  \right)}\implies \cfrac{sin(x)\cdot 0~~+~~cos(x)\cdot 1}{cos(x)\cdot 0~~-~~sin(x)\cdot 1}&#10;\\\\\\&#10;\cfrac{cos(x)}{-sin(x)}\implies -cot(x)
7 0
3 years ago
6+9=?<img src="https://tex.z-dn.net/?f=6x%5E%7B2%7D%20%2B7x-9%3D" id="TexFormula1" title="6x^{2} +7x-9=" alt="6x^{2} +7x-9=" ali
ELEN [110]

Answer:

are u sure thet write the question right?

4 0
3 years ago
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