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vladimir1956 [14]

3 years ago

9

I need help with filling in this table, please help!!! I will give brainiest to whoever answers correctly!!!

1 answer:

Sphinxa [80]3 years ago

3 0

Answer:

-4, 14+(-5) (-0.8), 14+ (3) (-0.8), 16.4

Step-by-step explanation:

The change in temperature for -5 is -4

The expression for -5 is 14+ (-5) (-0.8)

The expression for 3 is 14+ (3) (-0.8)

The actual temperature for 3 is 16.4

I did the math.

You might be interested in

Determine the equation of the following polyromial that passes through point (0,6).

Ostrovityanka [42]

Answer:

Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).

So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2

f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)

f(0)=−6f(0)=−6

a(1)(−2)=−6a(1)(−2)=−6

a=3a=3

f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)

f(x)=3x3−9x−6

3 0

3 years ago

A lamp post with a height of 12 ft casts a shadow of 6 ft. A fire hydrant near it casts a shadow that is 1.5 ft long. How tall i

alex41 [277]

Answer:

3ft

Step-by-step explanation:

We are given that

Height of lamp, h=12ft

Height of shadow, l=6ft

Height of shadow of hydrant, l'=1.5ft

Let height of hydrant=h'

We have to find the height of fire hydrant.

All right triangles are similar

When two triangles are similar then, the ratio of their corresponding sides are equal.

Using the property

$\frac{h}{h'}=\frac{l}{l'}$

$\frac{12}{h'}=\frac{6}{1.5}$

$h'=\frac{12\times 1.5}{6}$

$h'=3$ ft

Hence, the height of fire hydrant=3 ft

3 0

3 years ago

What is the slope of a line that is parallel to the graph of 5x+4y=y?

torisob [31]

Answer:

-5/3

Step-by-step explanation:

7 0

3 years ago

Add. –32.7+(–0.2)<br> a. –32.9<br> b. –32.5<br> c. 32.5<br> d. 32.9

yKpoI14uk [10]

The answer to your question is -32.9

5 0

3 years ago

Ayuda con mates por favor

QveST [7]

Answer:

a) $x = -1$ and $y = 5$ , b) $x = 2$ and $y = -1$ , c) $x = -2$ and $y = 1$ , d) $x = 2$ and $y = -9$ , e) $x = 2$ and $y = -2$ , f) $x = 2$ and $y = -1$ , g) $x = 2$ and $y = -1$ , h) $x = 1$ and $y = 2$ .

Step-by-step explanation:

Each system is solved as follows (Cada sistema es resuelto como sigue):

a) $2\cdot x + y = 3$ and $3\cdot x - y = -8$

$y = 3 - 2\cdot x$

$3\cdot x - 3 + 2\cdot x = -8$

$5\cdot x = -5$

$x = -1$

$y = 5$

b) $x - 2\cdot y = 4$ and $-x+3\cdot y = -5$

$x = 4 + 2\cdot y$

$-4-2\cdot y+3\cdot y = - 5$

$-4+y = -5$

$y = -1$

$x = 2$

c) $-2\cdot x + 5\cdot y = 9$ and $x - y = -3$

$x = y-3$

$-2\cdot y +6 +5\cdot y = 9$

$3\cdot y = 3$

$y = 1$

$x = -2$

d) $5\cdot x - 4\cdot y = 2$ and $3\cdot x + 2\cdot y = -12$

$3\cdot x + 2\cdot y = -12$

$6\cdot x + 4\cdot y = -24$

$4\cdot y = -6\cdot x -24$

$5\cdot x +6\cdot x +24 = 2$

$11\cdot x = -22$

$x = 2$

$y = -9$

e) $x + y = 0$ and $2\cdot x +3\cdot y = -2$

$y = -x$

$2\cdot x -3\cdot x = -2$

$-x = -2$

$x = 2$

$y = -2$

f) $3\cdot x + 5\cdot y = 1$ and $x + y = 1$

$y = 1 - x$

$3\cdot x + 5 - 5\cdot x = 1$

$-2\cdot x = -4$

$x = 2$

$y = -1$

g) $5\cdot x - 2\cdot y = 12$ and $4\cdot x + 3\cdot y = 5$

$x = \frac{12+2\cdot y}{5}$

$x = \frac{5-3\cdot y}{4}$

$\frac{12+2\cdot y}{5} = \frac{5-3\cdot y}{4}$

$4\cdot (12+2\cdot y) = 5\cdot (5-3\cdot y)$

$48 + 8\cdot y = 25 - 15\cdot y$

$23\cdot y = -23$

$y = -1$

$x = 2$

h) $7\cdot x + 8\cdot y = 23$ and $3\cdot x + 2\cdot y = 7$

$3\cdot x + 2\cdot y = 7$

$12\cdot x + 8\cdot y = 28$

$8\cdot y = 28 - 12\cdot x$

$7\cdot x +28-12\cdot x = 23$

$-5\cdot x = -5$

$x = 1$

$y = 2$

8 0

3 years ago