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Daniel [21]
3 years ago
15

If the temperature is 30°C, what is it in Fahrenheit?

Mathematics
2 answers:
julia-pushkina [17]3 years ago
6 0

Answer:

86.0 degrees fahrenheit

miss Akunina [59]3 years ago
5 0

Answer:86

Step-by-step explanation:

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Could someone by any chance help me with this problem?
Oksi-84 [34.3K]

Answer:

D

Step-by-step explanation:

Over 6 is the blue and red bars

the blue bar has 4 people

the red bar has 2 people

2+4=6

D

6 0
3 years ago
Read 2 more answers
Veronica wants to check her work after evaluating . What steps can she follow to verify her answer?
butalik [34]

Answer:

Divide her answer by -6 and see if her result is -108

Step-by-step explanation:

x ÷ -6 = -108

if not then the answer would be wrong

3 0
3 years ago
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Evaluate c−d if c=1 and2/5 and d=5/6 . Express your answer in simplest form.
Andrews [41]

Answer:

\frac{17}{30}

Step-by-step explanation:

→ Substitute in the numbers

1\frac{2}{5} -\frac{5}{6}

→ Simplify

1\frac{2}{5} -\frac{5}{6}=\frac{7}{5} -\frac{5}{6}=\frac{42}{30} -\frac{25}{30} =\frac{17}{30}

5 0
2 years ago
Can you please answer this question
tamaranim1 [39]
3tan^{2} \theta +7sec\theta=3
First I converted the equation terms into sine and cosine.
tan^{2}\theta = \frac{sin^{2}\theta}{cos^{2}\theta} and sec\theta= \frac{1}{cos\theta}
Substitution:
\frac{3sin^2\theta}{cos^2\theta} + \frac{7}{cos\theta} =3
Common Denominator Created:
\frac{3sin^2\theta}{cos^2\theta} + \frac{7cos\theta}{cos^2\theta} =3
Multiply each term by the LCD:
3sin^2\theta+7cos\theta=3cos^2\theta
Substitution: Recall ⇒sin^2\theta =1-cos^2\theta 
3(1-cos^2\theta)+7cos\theta=3cos^2\theta
Distribute and collect all terms on one side:
6cos^2\theta-7cos\theta-3=0
Factor and set each factor equal to 0:
(2cos\theta-3)(3cos\theta+1)=0
2cos\theta-3=0⇒theta=cos^{-1} \frac{3}{2}
3cos\theta+1=0⇒theta=cos^{-1} \frac{-1}{3}
The 2nd factor provides only possible answer 109.5 degrees

4 0
3 years ago
Help me here i suck at algebra due tomorrow plz explain step by step my english is not very good.
Hoochie [10]
Jhshegdgdvdvdfdfefvsg

5 0
3 years ago
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