निम्नलिखित जानकारी के आधार पर 50 से 60 शब्दों में विज्ञापन तैयार कीजिए :
निम्नलिखित जानकारी के आधार पर 50 से 60 शब्दों में विज्ञापन तैयार कीजिए :
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Solution and Explanation
विज्ञापन :
कोविड टीकाकरण केंद्र
स्वास्थ्य सुरक्षा के लिए टीका लगवाएँ!
कोविड-19 महामारी से बचाव के लिए कोविड टीकाकरण केंद्र में दिनांक, समय, स्थान और संपर्क जानकारी के साथ टीकाकरण की सुविधा उपलब्ध है। हम आपको सुरक्षित रखने के लिए आयु मापदंड के आधार पर टीके की डोज़ प्रदान करते हैं।
संपर्क करें:
टीकाकरण के लिए आज ही संपर्क करें और सुरक्षित रहें।
Questions Asked in Maharashtra Class X Board exam
- Appreciation of the poem:
Read the following poem and write an appreciation of it with the help of the points given below:
Stopping by Woods on a Snowy Evening
Whose woods these are I think I know.
His house is in the village, though;
He will not see me stopping here
To watch his woods fill up with snow
My little horse must think it queer
To stop without a farmhouse near
Between the woods and frozen lake
The darkest evening of the year.
He gives his harness bells a shake
To ask if there is some mistake.
The only other sound’s the sweep
Of easy wind and downy flake.
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
Robert Frost
- Maharashtra Class X Board - 2025
- Reading Comprehension
- Pick out the infinitive from the following sentence :
He was asking to go for the concert.- Maharashtra Class X Board - 2025
- Grammar and Language Usage
- Out of the following which is a Pythagorean triplet ?
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
- Maharashtra Class X Board - 2025
- Properties of Triangles
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.