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3 years ago

A recipe that makes 8 jumbo blueberry muffins calls for 1 1/2

2 answers:

6 0

A dozen is 12 jumbo muffins, so 3 dozen is 36 muffins. We Know how much sugar is needed for 8bis equal to 4. So we need 4Xs the previous amount. 1.5 tspx4=6 teaspoons

artcher [175]3 years ago

5 0

The dozen is 36...36/8= 4.5. So 1.5x4.5=6.75 which converts to 6 3/4 teaspoons

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Find the Area Of each circle.

Levart [38]

The area of the circle A and B is 86.54 inches and 124.62 mm

According to the statement

we have given that the radius of the circle A and b and we have to find the area of the given circle.

So, we know that the

The area enclosed by a circle of radius r is πr².

So, For area of the circle

For condition A :

diameter = 10.5 inches

then radius = 5.25

Area = πr²

Area = 3.14*27.56

Area = 86.54 inches

Now, For condition B :

radius = 6.3 mm

Area = πr²

Area = 3.14*39.69

Area = 124.62mm

So, The area of the circle A and B is 86.54 inches and 124.62 mm

Learn more about area of the circle here

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1 year ago

(English to algebra) twice the difference of w less than five

denpristay [2]

i believe its 2(5-w)

5 0

2 years ago

Read 2 more answers

(16 divided by 4) x (7-4)

zavuch27 [327]

4 x 3 = 12
12 would be the answer

4 0

3 years ago

Read 2 more answers

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .