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3 years ago

I need help with this answer

Mathematics

1 answer:

astra-53 [7]3 years ago

3 0

Answer:

Yes, it is a linear expression because the highest power of x is 1.

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If you could help me on all of them you a live saver

PtichkaEL [24]

Answer:

(1.37) AUB = { 1,2,3,4,5,6}

(1.38) AUC = { 1,2,3,4,5 }

(1.39)BUC = { 1,2,3,4,5,6}

(1.40) { 2,4 }

(1.41) { 1,3,5 }

(1.42) { phi }

(1.43) AU(BUC) = { 1,2,3,4,5,6 }

(1.44) { phi }

(1.45) {1,2,3,4,5}

(1.46) { 1,2,3,4,5 }

$\huge\underline\red{Hope\:that\:helps}$

8 0

3 years ago

Yesterday,Mr.Swenson made 2 3/4 quarts of strawberry jam and 1 1/8 quarts of raspberry jam. How much more strawberry jam did he

Ivenika [448]

Step-by-step explanation:

So I found the l.c.m which is 8 and I multiplied it by the numbers and then

The subtracted the both mixed numbers and I got an answer of 15/8

3 0

2 years ago

30 POINTS I WILL GIVE U BRAINLIEST AND THANKS AND 5 STARS please help this is very important this is a big part of my grade

mars1129 [50]

Answer:

$1.\:2^x=64, x=\boxed{6},\\2.\:x=\left(\frac{2}{3}\right)^3,x=\boxed{\frac{8}{125}}\\3.\:3\cdot (3^4)=3^x, x=\boxed{5}\\4.\:\frac{16}{25}=x^2,x=\boxed{\frac{4}{5}},$

Step-by-step explanation:

$1.\\\\2^x=64,\\\log 2^x=\log64,\\x\log 2=\log 64,\\x=\frac{\log 64}{\log 2}=\boxed{6}$

$2.\\x=\left(\frac{2}{5}\right)^3=\frac{2^3}{5^3}=\boxed{\frac{8}{125}}$

3. This problem incorporates an exponent property.

Exponent property used: $a^b\cdot a^c=a^{(b+c)}$ , yielding an answer of $\boxed{5}$

$4.\\\\\frac{16}{25}=x^2,\\x=\sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\boxed{\frac{4}{5}}$

3 0

3 years ago

Read 2 more answers

What is the slope of this graph?

sleet_krkn [62]

Answer:

-3

Step-by-step explanation:

The graph goes down 3 for each one on the x-axis

6 0

2 years ago

Read 2 more answers

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .

7 0

2 years ago