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3 years ago

I have 2 mor quizzes of this assiment day 1 of trying to start 9th grade language arts

Mathematics

1 answer:

Scilla [17]3 years ago

3 0

The change in temperature is -28°F

So, Option D is correct.

Step-by-step explanation:

Temperature at day time = 72°F

Temperature at night = 44°F

What number represents the change in number?

Change in temperature = Temperature at night - Temperature at Day

Change in temperature = 44°F-75°F

Change in temperature = -28°F

The change in temperature is -28°F

So, Option D is correct.

Keywords: Solving Expressions

Learn more about Solving Expressions at:

#learnwithBrainly

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Caitlin has a $10 gift certificate to the music store. she has chosen a number of cd's from the $7 bargain bin. if the cost of t

Tcecarenko [31]

She bought x number of $7 cd and used and $10 gift card so
7x-10=32
7x=42
x=6 cds

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3 years ago

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Seven tenths of a number plus 14 is less than forty-nine

Virty [35]

7/10n + 14 = 49
7/10n = 35
n = 35 x 10/7
n = 50

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3 years ago

J u l i a n has partially filled the p r i s m below with 16 unit cubes. How many more unit cubes will he need to completely fil

Wittaler [7]

add it

Step-by-step explanation:

11+20+25+36

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago

Find the area of rectangle $BCEF$ . Round the area to the nearest whole number, if necessary.

WINSTONCH [101]

Answer:

999

Step-by-step explanation:

5 0

3 years ago